An Introduction to Matrix Convex Sets and Free Spectrahedra

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We give a new and very simple proof of the afore-mentioned weak Krein-Milman • theorem by Webster and Winkler and show that the closure is superfluous making it a weak Minkowski theorem (Theorem 6.8). The Gleichstellensatz characterizes the smallest monic linear (matrix) pencil L = • I L1X1 ... LgXg defining a given free spectrahedron S = L := A g− − − D { ∈ S I I L1 A1 ... Lg Ag 0 . In fact, this smallest pencil is a direct summand| ⊗ − of every⊗ other− monic− linear⊗ pencil} defining the same spectrahedron. The original proof of Helton, Klep and McCullough uses the concept of the Silov ideal, a complicated object from the theory of operator systems and C∗-algebras. We present two new more elementary proofs of the Gleichstellensatz (Corollary 5.17 and Corollary 6.13). We introduce a new technique to analyze matrix convex sets of the form S = A • { ∈ Sg t [0, 1] : p(tA) 0 , where p is a noncommutative polynomial. This technique| ∀ ∈ allows us to give≻ a} new variant of the proof of a theorem of Helton and McCullough stating that S is the interior of a free spectrahedron (Theorem 4.2). Another application is that we are able to infer that spectrahedra can be written as intersections of irreducible spectrahedra. We prove that for a monic linear pencil L either L is reducible or the sequence (detk L)k∈N of determinants of L restricted to the k-th level set Sg(k) become irreducible for big k (Corollary 5.24).

1.2. Motivation. A spectrahedron is a set of the form g g (x1, ..., xg R n N, L0 + Lixi 0 ( }∈ ∈ ) i=1 X where the L are Hermitian matrices of a uniform size. It has turned out that spectrahedra i are suitable for numerical calculations, especially because they are convex. For the problem of optimizing a linear functional over a spectrahedron (this task is called semidefinite programming and generalizes linear programming) there exist efficient interior-point algo- rithms. Therefore in recent years much effort has been made to approximate polynomial optimization problems by semidefinite programs or hierarchies of those (e.g. the θ-number of a graph, Lasserre relaxation; see [BV] and [L]). In Engineering, especially in linear systems and control theory, so-called matrix inequalities appear in pratical problems. The most common approach to solve them is to try to find an equivalent formulation in terms of a semidefinite program (see [BEFB] and [HKM2]). For those reasons many papers dealing with the question which sets are spectrahedra or at least projections of spectrahedra have been written (see [HN1], [HN2], [KS]). Another area of research connected to spectrahedra is the question which hyperbolic polynomials admit a determinantal representation (see [HV]). Consider now the following linear matrix inequality (LMI) g

L I + Li Xi 0 0 ⊗ ⊗ i X=1 2 (the left is side is also called a (linear) pencil and in case that L0 = I this pencil is called monic). This expression can be evaluated in the Xi with Hermitian matrices of an arbitrary common size. The solution set g k×k g (X1, ..., Xg (SC ) L0 Ik + Li Xi 0 , (I) ( }∈ ⊗ ⊗ )! i=1 k∈N X where denotes the Kronecker tensor product and SCk×k the Hermitian k k-matrices, is called⊗ a free spectrahedron and fulfills some stronger convexity property, which× is called matrix convexity. Matrix convex sets C are closed under matrix convex combinations, i.e. r r ∗ ∗ A , ..., Ar C, V Vj = I = V AjVj C, 1 ∈ j ⇒ j ∈ j i X=1 X=1 ∗ ∗ where the Vj are rectangular matrices of appropriate sizes and V AjVj := (V Aj Vj i j j i | ∈ 1, ..., g ). Note that the Aj do not have to be all of the same size. { } We see that if we specialize our solution set (I) to vectors (tuples of Hermitian 1 1- matrices), we end up with the (ordinary) spectrahedron from the beginning. Given× a spectrahedron defined by a pencil, one can hope that key properties of its geometry are reflected by a particularly nice form of the pencil. Here it turns out that the right framework for following this purpose is the free setting. An example is for instance the following question: Given two spectrahedra with monic defining pencils L1 and L2, when is the first one contained in the second one? No sharp characterisation for this property is known. A straight-forward certificate for containment would be a sums of squares-representation of L2 in terms of L1. However this would imply also the containment of the related free spectrahedra. Indeed, for free spectrahedra such a sums-of-squares representation exists if and only if the free spectrahedron given by L1 is contained in the one given by L2 [HKM1, Lemma 4.7]. Morally speaking, a spectrahedron can be defined by many different pencils. Evaluating those pencils in tuples of Hermitian matrices and not only vectors makes this hidden structure visible. Therefore a big motivation behind the study of free spectrahedra is to find out more about (ordinary) spectrahedra. However the free theory still needs to evolve in order to provide powerful techniques for the study of ordinary spectrahedra. One example of a succesful result approached in this way is the main result of [HKMS], which says that one monic pencil L2 defining a much bigger spectrahedron than a monic pencil L1 defines automatically a bigger free spectrahedron than the pencil L1 and hence L2 has a sums-of-squares representation with L1 as a weight. Another source of motivation to study matrix convex sets is its connection to the theory of operator systems and completely positive maps. The structure of an operator system is determined up to complete isometry by its matrix range, which is a compact matrix convex set. A further instance where matrix convexity is utilized is the theoretical study of so- called dimension independent polynomial matrix inequalities. Some problems in the theory of linear systems and control are equivalent to solving a polynomial matrix inequality p(A, Y ) 0 given by a noncommutative polynomial p with a fixed parameter Y . The parameter Y determines the size of the solution matrices A. Considering those problems for fixed parameters and transforming them to a problem over commutative variables is not well-behaved in the sense that two different parameters lead to very different and seemingly unrelated problems. The idea is now to stick to the non-commutative setting where the the problem keeps the same structure for different choices of parameters (one says that the problem scales automatically when changing dimensions because noncommutative 3 polynomials behave well with respect to direct sums and compressions). In case that the polynomial matrix inequality defines a convex problem (or at least convex in the variables A) the theory of (partial) matrix convexity can be applied (see [HHLM] and [HMPV]). The research of matrix convex sets started in the early 1980s. First fundamental results were given in the 1990s, however the number of researchers working in this field was relatively small. Nowadays the subject is increasing and drawing more attention. Matrix convexity has a strong connection to operator theory, in particular to completely positive maps. Other related topics are free real algebraic geometry and free complex analysis. One aim of this work is to present a unified introduction to matrix convexity and free spectrahedra. For non-specialists, who want to know more about the topic, there are some obstacles we want to reduce. Some important theorems are not easy to understand because the proofs use advanced results and concepts from operator theory (e.g. Gleich- • stellensatz [HKM4]) or the theorem is stated in too much generality resulting in a more involved proof and • weaker conclusion (e.g. weak free Krein-Milman theorem [WW]) or the proof is long (e.g. Helton-McCulloughs characterization of free spectrahedra as • closures of matrix convex free basic open semialgebraic sets [HM]). 1.3. Notation and basic definitions.

Notation 1.1. For this article let g N and ﬁx a tuple X = (X1, ..., Xg) of g free noncommutative variables. The letter ∈i will be exclusively denote elements of 1, ..., g . In this paper all rings shall contain 1 and ring homomorphisms are required to{ be unital} except of cases where we explicitely use the word "non-unital". If R is a commutative ring, then a polynomial over R in the variables X is the formal object

f = fαXα(1)...Xα(d) d dX∈N0 α∈{X1,...,g} given by a coefficients fα R where only finitely many fα are non-trivial. The degree of ∈ d f = 0 is the maximal d such that there is α 1, ..., g with fα = 0. Two polynomials are6 equal if and only if all coefficients are equal.∈ { Scalars} from R shall6 commute with the variables X1, ..., Xg, i.e. rXi = Xir for i 1, ..., g and r R. We form the sum and product of two polynomials in the obvious way∈ { and denote} the∈ polynomials in the variables X over R by R X (NC polynomials). This object is also called the free R-algebra over g generators.h Iti has the universal property that for every not necessarily commutative R-algebra B which contains R in its center and b , ..., bg B there is a unique R-algebra 1 ∈ homomorphism ϕ : R X B such that ϕ(Xi) = bi for i 1, ..., g . Let g be the h i → ∈ { } F free monoid generated by g elements X , ..., Xg. One can identify g as the monomials of 1 F C X . The free (complex) algebra C X admits an involution ∗. For p C X we get p∗ byh conjugationi of the coefficients andh reversingi the order of the multiplication∈ h ofi variables in each monomial.

Contrary to the noncommutative case, if (Y1, ..., Yr) are commuting variables, we write R[Y1, ..., Yr] for the polynomial ring over R. If T is a ring with a notion of degree and d N, then Td shall denote the elements of T with degree at most d. We denote with ∈ SCk×k the Hermitian k k matrices with entries from C. × g g k×k g Much of the work will take place in the free space S := (S (k))k∈N := ((SC ) )k∈N, the set consisting of the collection of g-tuples of Hermitian square matrices of uniform size. The elements of this set will serve as point evaluations for the noncommutative polynomial ring. Sg(k) will be called the k-th level (set) of the free space. For A Sg(k) and 4 ∈ k×r ∗ ∗ ∗ g g 2 B C we set B AB = (B A B, ..., B AgB) S (r). We define A = A ∈ 1 ∈ || || || i=1 i || where . denotes the operator norm. q || || P δ×ε Consider now a noncommutative matrix polynomial f (C X ) for some ε, δ N0. This is nothing else than a matrix with polynomial entries.∈ h Eiquivalently one can∈ also interpret it as a polynomial with matrices as coefficients. The maximum of the degrees of the entries of f will be the degree of f. f ∗ is obtained by transposing f and afterwards applying the involution on C X to each entry. Write f = α∈F g fαα with matrices δ×ε h i k×k g kδ×kε fα C . For a square matrix A (C ) we define f(A) C to be f(A) = ∈ ∈ P∈ α∈F g fα α(A) where the evaluation α(A) is obtained by replacing every occurance of ⊗ 0 Xi in α by Ai. Here denotes the Kronecker product of matrices and A = Ik the k k- Pidentity matrix. We remind⊗ the reader of the calculation rule (a b)(c d)= ac bd×for matrices of appropriate sizes. Another way of defining f(A) is to⊗ consider⊗ f as a⊗ matrix ∗ g with polynomial entries and substitute Ai for Xi in each entry. If f = f and A S (k), ∈ then f(A) SCδk×δk. ∈ All Hilbert spaces in this paper shall be separable and complex. If , are such Hilbert spaces, we write H K

( , )= A : linear and bounded , B H K { H → K } ∗ ( ) = ( , ) and h( ) = A ( ) A = A . For A ( ) we define f(A) = B H B H H B H { ∈ B H | } ∈ B H g fα α(A) in the above way as well (we will also allow that the coefficients fα α∈F ⊗ take values in some ( )). We write (∞) := , which becomes a Hilbert space P n∈N B K H2 H (∞) (∞) by setting an = an . For B ( ) we obtain B ( ) by || n∈N || n∈N || || L∈ B H ∈ B H declaring B(∞) a = Ba . L n∈N n pP n∈N n g δ×δ For L S (δ) andL C SC L , we define C LX = C (L1X1 + ... + LgXg). This expression∈ is a Hermitian∈ matrix polynomial of− degree 1 and− size δ. Matrix polynomials of these form are called (linear) pencils (even though one could make an argument that they should be called affine linear). We say that the pencil is monic if C = I. We adopt the following convention: For L Sg(δ), we denote by L the monic pencil I LX. If on ∈ − the other hand L is a monic linear pencil, then L = ( L(e1) I, ..., L(eg) I) will be its truly linear part (in case we work with two monic− pencils− we will− also use− the letters H and H). A linear pencil C BX defines an associated (free closed) spectrahedron = − DC−BX A Sg (C BX)(A) 0 where (C BX)(A) 0 means that (C BX)(A) is positive{ ∈ semidefinite| − ( stands } for "positive− definite"). For a separable Hilbert− space g ≻ g H and B h( ) ,C h( ) we also set C−BX = A S (C BX)(A) 0 , however such a∈B set willH not be∈B a freeH spectrahedronD in general.{ ∈ We| will −focus our analysis } of free spectrahedra on those which have an interior point x (1) in Rg or equivalently ∈ DC−BX admit a description with a pencil C BX which is positive definite in this point. Most of the time we will assume that this point− is zero. Then one can find a description of this spectrahedron by a monic linear pencil L (Proposition 1.12). Notice that the first level (1) coincides with the ordinary spectrahedron defined by DB−CX B CX in Rg. All other levels (k) are also ordinary spectrahedra in Sg(k). For − DB−CX a monic pencil L the set L(k) can be also determined as the closure of the connected D g component of the set A S (k) detk L(A) = 0 around 0 where detk L denotes the g{ ∈ | 6 } determinant detk L : S (k) R, A det(L(A)). → 7→ 5 k ∗ Let k N. For α 1, ..., k let eα C be the α-th unit vector and set Eα,α = eαeα. For

∈ ∈ { } ∈ ∗ ∗ ∗ ∗ i α, β 1, ..., k with α < β set Eα,β = eαe + eβe and Eβ,α = ieαe eβe . We deﬁne ∈ { } β α β − α the g-tuple of generic k k matrices as × k 1 = Eα,β α,β X X α,βX=1 1 g where the α,β = ( , ..., ) are g-tuples of variables and the tuple X Xα,β Xα,β ( i ) Xα,β i∈{1,...,g},α,β∈{1,...,k} consists of commuting variables. We surpress the dimension k in this notation.

For a monic linear pencil L the determinant detk L( ) is an example of a real-zero X polynomial (RZ-polynomial). These are defined to be polynomials p R[Y , ..., Yn] such ∈ 1 that p(0) = 0 and for all y Rn 0 the univariate polynomial p(Ty) has only real roots. 6 ∈ \ { } The connected component of p−1(R 0 ) around 0 is automatically convex. We refer the reader to [HV] for more material on\ RZ-polynomials. { } g δk δ×δ Let A S (k), v C and C BX S(C X )1 be a linear pencil with (C BX)(A)v = ∈ δ∈ − ∈ k h i − 0. Write v = eα vα with vα C . Set M(A, v) = span(v , ..., vδ ). α=1 ⊗ ∈ C−BX 1 Free spectrahedraP are an important example of matrix convex sets (Proposition 1.3). A set S Sg is called matrix convex if ⊆ r r kj×s ∗ ∗ Aj S(kj),Vj C , V Vj = Is = V AjVj S(s) ∈ ∈ j ⇒ j ∈ j j X=1 X=1 ∗ ∗ ∗ We remind the reader that Vj AjVj is defined as (Vj (Aj)1Vj, ..., Vj (Aj)gVj). For a unitary matrix U Ck×k and A S(k) this implies that also U ∗AU S(k). Therefore matrix ∈ ∈ ∈ convex sets are closed under unitary conjugation. For A, B Sg we will write A B if there is a unitary matrix U such that A = U ∗BU. Obviously,∈ each level S(k) ≈of a matrix convex set S is convex. Matrix convex sets are also closed under forming direct sums meaning that A, B S implies A B := (A B , ..., Ag Bg) S. If A S(k) ∈ ⊕ 1 ⊕ 1 ⊕ ∈ ∈ and P : Ck im(P ) Ck is an orthogonal projection, then P P ∗ = I and thus P AP ∗ S(rk(P )). In→ this paper⊆ projections are treated as surjective maps; if we want to view them∈ as endomorphisms, we write P instead of P . If 0 S, then matrix convexity can be restated as ∈ r r kj ×s ∗ ∗ Aj S(kj ),Vj C , V Vj Is = V AjVj S(s). ∈ ∈ j ⇒ j ∈ j j X=1 X=1 For S Sg the set mconv(S) := T Sg T matrix convex , T S is called the ⊆ { ⊆ | ⊇ } matrix convex hull of S and is the smallest matrix convex superset of S in Sg. T There are reasons why we chose our matrix convex sets to be subsets of Sg; however, in principle it would be also natural to allow that g-tuples of Hermitian operators on an infinite-dimensional separable Hilbert space are elements of our free space. Both view points have their advantages (The finite-dimensional setting leans itself more to practical computations and the theory of ordinary spectrahedra; if one allows ordinary spectrahedra to be defined by LOI (Linear operator inequality), then every closed convex set would be

1 Our definition of the matrix units Eα,β is non-standard. We adapted it in order to be suited for the ∗ study of hermitian matrix tuples. The normal matrix units eαeβ are not all hermitian. 6 a spectrahedron defined by a diagonal LOI. For either setting there are theorems which have cleaner statements than in the other setting.). The following notation is reminiscent of this hidden additional structure, which becomes important in some situations. If is a g H separable Hilbert space and L h( ) , then we set ∈B H mconv(L):= mconv P LP ∗ P : im(P ) is a finite-dimensional projection Sg { | H→ }⊆ g where the closure of a set S S is defined as S = (S(k))k∈N. (We will prove later that for finite-dimensional⊆ the closure in the definition of mconv(L) is superfluous (CorollaryH 1.9), making this definition consistent with the previous one.) We have seen that matrix convex sets are invariant under unitary conjugation. Therefore when dealing with matrices, we treat them sometimes rather as linear operators than representations of operators with a fixed chosen orthonormal basis. For example we call a matrix A a submatrix (resp. direct summand) of a matrix B if there is a unitary matrix U and matrices A C A 0 C,D,E such that U ∗BU = resp. U ∗BU = . D E 0 E For k N we denote by k−1 Ck the (k 1)-dimensional sphere. If (U, . ) is a normed ∈ S ⊆ − || || vector space, a U, r > 0, then BU (a, r)= x U x a < r is the ball of radius r around a. Sometimes∈ we omit the ambient space{ ∈U in| || this− notation.|| } 1.4. Content of the paper and readers guide. Overview 1.2. In the rest of the introduction we will cite well-known results about matrix convex sets and free spectrahedra, which are needed for the further exposition. Our methods incorporate techniques from real algebraic geometry, C∗-algebras and the theory of completely positive maps. Since most readers will be familiar with the first two branches of mathematics, we will mention quickly what we need from there in the appendix (real closed fields, Tarski transfer principle, Finiteness theorem for semialgebraic classes, in- finitesimals and standard part; characterization of C∗-algebras as closed subalgebras of the algebra of bounded linear operators on a Hilbert space, basic representation theory, Burnsides theorem). The theory of completely positive maps is helpful for the analysis of matrix convex sets and conceptually very nice, however not completely necessary. In most cases one can get away with the Effros-Winkler separation technique (Chapter 2.1) as a replacement. Some papers combine these two concepts, even though one is enough for most purposes. Therefore complete positivity will be further explained in the appendix and it is up to the reader to read Chapter 2.1 or the appendix or both. As standard references we give [BCR] (real algebraic geometry), [D1] (C∗-algebras) and [P] (completely positive maps). Similar to the theory of ordinary convexity, separation techniques are essential and valuable for the analysis of matrix convex sets. The Effros-Winkler separation will be explained in Chapter 2. We will generalize it in order to also handle non-closed sets (Corollary 2.17; in order to expose matrix extreme points) and even situations where only weak separation is possible (Corollary X.34 in the appendix). For the latter purpose we will allow separating linear forms to attain values in a real closed extension field of R.

In Chapter 3 we will assign two numbers pz(S), kz(S) N0 to each matrix convex set S. These numbers will indicate whether the whole∈ S is∪ generated {∞} by one level S(k) in a maximal/minimal fashion. We analyze how these numbers are aﬀected by taking the polar of S (Corollary 3.7). As an outcome we will see how the pz-number of a free spectrahedron encodes how small the maximal size of the blocks occuring in the block diagonalization of a pencil description of the given free spectrahedron can be (Proposition 3.8 7 and Corollary 3.10). In [HM] Helton and McCullough characterized free spectrahedra as "closures of matrix convex free basic open semialgebraic" sets. A new proof and slight generalization of Hel- ton and McCulloughs theorem is given in Chapter 4. The employed techniques will also prove to be fruitful in later chapters. The main results are Theorem 4.2 and Corollary 4.11.

In Chapter 5 we will see that every free spectrahedron L can be expressed as an intersec- D tion of "irreducible" spectrahedra L . Under a natural minimality condition, the pencils D j Lj will be uniquely determined (up to unitary equivalence). This will enable us to decompose the pencil L into a direct sum of pencils involving all the Lj (Corollary 5.17). As a consequence we obtain a new/simplified proof of the Gleichstellensatz [HKM4, Theorem 3.12] and its generalization [Z, Theorem 1.2]. Furthermore we will consider a monic linear pencil L and study the system of determinants detk L( ) as polynomials on Hermitian k k-matrices (k N). We will see how a X × ∈ decomposition of detk L in factors in each level k gives a decomposition of the free locus (L) = X Sg ker L(X) = 0 as a union of "irreducible loci" (Corollary 5.26). ThisZ will{ lead∈ to a| characterization6 { }} of pencils which are "irreducible" and "minimal" (Theorem 5.20). One can define different notions of extreme points of matrix convex sets. These will be investigated in Chapter 6. We will give a new proof of the weak free Krein-Milman theorem from [WW] for compact matrix convex sets S in Sg and generalize it to a free Minkowski theorem (Theorem 6.8). For sets S with finite kz-number we will strengthen the theorem. We prove that the matrix convex hull of the absolute extreme points of S equals S (Corollary 6.12). This culminates in another proof of the classical Gleichstel- lensatz (Corollary 6.13). Additionally we introduce matrix exposed points, characterize them (Theorem 6.20) and prove a free Straszewicz theorem (Corollary 6.22). Furthermore we give examples that compact matrix convex sets do not need to have absolute extreme points (Example 6.28). Finally we prove a general Gleichstellensatz/strong free Krein- Milman theorem for compact matrix convex sets which shows that every such set S admits a smallest description as the matrix convex hull of an operator tuple L (Theorem 6.42). In the tuple L the absolute extreme points of S and a generalized form of absolute extreme points are encoded. Afterwards we will take the chance to pause and give some examples of the applications of the accumulated theory up to Chapter 7. References to the examples are given in the earlier chapters. However some techniques are introduced only after the references. The purpose of Chapter 8 is to analyze the connection between the sequence of determinants (detk L( ))k∈N of a linear pencil L and pz( L). We make some beginning steps towards answeringX the question when a sequence ofD RZ polynomials comes from the determinant of a pencil.

Finally, Chapter 9 contains an analysis of the degrees of the determinants detk L of a monic linear pencil L. We show that there is N N and ε > 0 such that deg(detk L) = kε for k N (Corollary 9.7) and that this N cannot∈ be set as 1 in general (Corollary 9.8). ≥ At the end, we have included an index for the notation.

1.5. More facts on matrix convex sets. In this section we collect some important facts and techniques for the study of matrix convex sets.

Proposition 1.3. Let B CX be a linear pencil. Then B−CX is matrix convex. − 8 D kj ×k Proof. Let r N, Aj B−CX (kj ) and Vj C for j 1, ..., r such that A := r ∗ ∈ ∈ Dr ∗ ∈ ∈ { } g j=1 Vj AjVj and I = j=1 Vj Vj. Then we have (B CX)(A) = B I + i=1 Ci r ∗ r ∗ g − ⊗r ∗ ⊗ ( V (Aj )iVj)= (I V )(B I + Ci (Aj)i)(I Vj)= (I V )(B P j=1 j jP=1 ⊗ j ⊗ i=1 ⊗ ⊗ j=1 ⊗P j − CX)(A)(I Vj) 0 P ⊗ P P P Definition 1.4. We say that a set S Sg is bounded/open/closed/compact if the respective level sets S(k) are bounded/open/closed/compact.⊆ A matrix convex set S is bounded if and only if S(1) is bounded (Proposition 1.5). We write int(S)(k) for the interior points of S(k). This hierarchy of sets constitutes int(S). Similarly we define the closure S. We say that S contains an open neighborhood of 0 if an ε> 0 exists such that g B g (0, ε)= A S A < ε S S { ∈ | || || }⊆ (for matrix convex S this is equivalent to the fact that 0 is an interior point of S(1) (Proposition 1.6)). We say that a subset T Sg is invariant under reducing subspaces if A B T implies A T . ⊆ ⊕ ∈ ∈ g Proposition 1.5. Let S S be matrix convex and r [0, ) such that S(1) B g (0, r). ⊆ ∈ ∞ ⊆ R Then S B g (0, r√g) is also bounded. ⊆ S Proof. Let A S(k) such that A 2 r2g. Choose i 1, ..., g , v k−1 such that 2 2 ∈ || || ≥ ∈ { ∗ }2 ∈ S2 ∗ Ai v r . WLOG v is an eigenvector of Ai. Then we have v Ai v r , v Av S(1) and|| ||v∗ ≥Av 2 r2. | | ≥ ∈ || || ≥ Proposition 1.6. [HKM1, Lemma 4.2] Let S Sg be matrix convex and r (0, ) such ⊆ g r ∈r ∞ that B g (0, r) S(1). Then S contains all A S with A (i.e. B g (0, ) S). R ⊆ ∈ || || ≤ g S g ⊆ g r Proof. [HKM1, Lemma 4.2] Let A S (δ) such that A g . Then each gAi r and δ×δ ∈ ∗ || || ≤ || || ≤ we can find Ci C unitary such that Ci gAiCi is diagonal and the diagonal entries ∈ ∗ are of norm r. Therefore (0, ..., 0,Ci gAiCi, 0, ..., 0) S. S is closed under unitary ≤ g ∈ 1 conjugation; thus (0, ..., 0, gAi , 0, ..., 0) S and A = (0, ..., 0, gAi, 0, ..., 0) S. ∈ i=1 g ∈ Lemma 1.7. Let S Sg be a matrix convex set andP ϕ : Sg R be affine linear such that ϕ(S(1)) = 0 . Then⊆ we can represent ϕ as the evaluation→ of a polynomial of degree { } 1 and are able to evaluate ϕ in points of Sg. We have ϕ(S)= 0 . { } Proof. Suppose that A S(k) and ϕ(A) = 0. Choose v k−1 such that v∗ϕ(A)v = 0. Then v∗Av S(1) and ϕ∈(v∗Av)= v∗ϕ(A)6 v = 0. ∈ S 6 ∈ 6 Lemma 1.8. (Free Caratheodory) Let A Sg(m), T Sg and A mconv(T ). Then ∈ ⊆ ∈ there exist C1, ..., C2gm2+1 T and matrices V1, ..., V2gm2 +1 of appropriate sizes such that 2 ∈ 2 2gm +1 ∗ 2gm +1 ∗ Im = j=1 Vj Vj and A = j=1 Vj CjVj.

d ∗ Proof.PWe find d N, C , ...,P Cd T and V , ..., Vd such that V Vj = Im and ∈ 1 ∈ 1 j=1 j d ∗ 2 ∗ ∗ j=1 Vj CjVj = A. Let d> 2gm + 1. We conclude that the matricesP (Vj CjVj Vj Vj) ∗ ∗ ⊕ − (V1 C1V1 V1 V1) (j 2, ..., d ) are R-linearly dependent. Choose λ = (λ2, ..., λd) Pd−1 ⊕ d ∈ { ∗ } ∗ d ∗ ∗ ∈ R 0 with j=2 λj(Vj CjVj V1 C1V1) = 0, j=2 λj(Vj Vj V1 V1) = 0 and set \ { }d − d − ∗ λ1 = j=2 λj.P Now we have for all α R thatPA = j=1(1 αλj)(Vj CjVj). For − d ∈ ∗ − α R smallP we can write A = j=1( 1 αλjVj Xj 1 PαλjVj). It is straightforward ∈ d ∗ − − that we still have ( 1 αλjV 1 αλjVj)= I. Now choose α R in such a way j=1 − P j p − p ∈ that all (1 αλ ) are nonnegative and one is zero. We have reduced one summand in the j P p p description− of A. 9 Corollary 1.9. [DDSS, Proposition 2.5] Let L Sg. Then mconv(L) is compact. ∈ The following is a nice characterization of matrix convex sets, which is seemingly easier to verify in practice because it does not feature complicated matrix convex combinations, which can be difficult to calculate with. Lemma 1.10. [HKM3, Lemma 2.3] Let S Sg. Then the following is equivalent: ⊆ (a) S is matrix convex. (b) For each level k N the set S(k) is convex. Furthermore S is invariant under unitary tranformations, taking∈ direct sums and reducing subspaces. Proof. [HKM3, Lemma 2.3] (a) = (b): This is easy. ⇒ (b) = (a): Suppose S fulfills (b) and A mconv(S)(δ). Then we find r N, Bj ⇒ k ×δ ∈ r ∗ ∈ ∈ S(kj) and matrices Vj C j for j 1, ..., r satisfying A = V BjVj and I = ∈ ∈ { } j=1 j V 1 P r ∗ r r . ∗ j=1 Vj Vj. We set k = j=1 kj, B = j=1 Bj, V =  .  and obtain A = V BV as Vr P P L   well as V ∗V = I. As S is closed under direct sums, we know B S. Extend V to a unitary ∈ W : Cδ Ck−δ Ck. Let P : Ck Cδ be the projection on the first δ components. Then ⊕ 7→ → A E A = P CP ∗ where C = W ∗BW S. Now write C = . ∈ E∗ F A 0 1 A E 1 1 0 A E 1 0 = + S 0 F 2 E∗ F 2 0 1 E∗ F 0 1 ∈ − − Here we used that S is convex and closed under unitary conjugation. Since S is also closed under taking reducing subspaces, we get A S. ∈ Remark 1.11. Let S Sg be a matrix convex set. Then the k-th level of S determines all lower levels. Indeed,⊆ fix A S(k) and let s < k. Choose an arbitrary projection P such ∈ that P AP ∗ S(k s). Let B Sg(s). Then we have B S(s) P AP ∗ B S(k). ∈ − ∈ ∈ ⇐⇒ ⊕ ∈ Proposition 1.12. [HKM4, Proposition 2.1] Let T HX S(C X )δ×δ be a linear pencil − ∈ h i 1 and S = A Sg(k) [T HX](A) 0 . Suppose that 0 is in the interior of S. Then { ∈ | − } there exists a monic linear pencil L of size δ such that L = S. D An important technique in the study of matrix convexity is to use projections to transform statements about a matrix convex set to statements about a certain level set. The following observation is one of the key properties of free spectrahedra. Lemma 1.13. (Projection lemma) [HKMS, Lemma 2.3] Let B SCδ×δ,C Sg(δ) and ∈ ∈ k N. Then (k) = A Sg(k) projections P : Ck im(P ) : dim(im(P )) ∈ DB−CX { ∈ | ∀ → ≤ δ = P AP ∗ . ⇒ ∈ DB−CX } g k Proof. [HKMS, Lemma 2.3] B−CX is matrix convex. Thus if A S (k) and P : C D ∗ g∈ → im(P ) is a projection, then P AP B−CX . Now let A S (k) and A / B−CX . δk ∗ ∈ D ∈ δ ∈ D Choose v such that v (B CX)(A)v < 0 and write v = α=1 eα vα with k ∈ S k − ⊗ vα C . Let P : C im(P ) be the projection onto span v , ..., vδ . Then we calculate ∈ → { 1 }P v∗(B CX)(P AP ∗)v = [(I P ∗)v]∗(B CX)(A)(I P ∗)v = v∗(B CX)(A)v < 0. − ⊗ − ⊗ − 2. Separation techniques for matrix convex sets An important technique in classical convexity consists in the separation of a point from a convex set that does not not contain that point. The following theorem transfers the 10 Hahn-Banach separation theorem (bipolar theorem) to the matrix convex setting. Version (a) was originally stated for bounded sets S however the assumption is not used in the proof. It shows that the role of linear functionals in the classical convex setting in Rg can be taken over by linear matrix inequalities in the free setting. Lemma 2.1. [HM, Proposition 6.4], [EW, Theorem 5.4] (Effros-Winkler theorem, separation of matrix convex sets and points via pencils) (a) Let S Sg be a closed matrix convex set with 0 S. Let Y / S. Then we can find a monic⊆ linear pencil L of the same size of Y which∈ separates∈S from Y in the sense that L(A) 0 for all A S and L(Y ) is not positive semidefinite. ∈ (b) Let S Sg be an open matrix convex set with 0 S. Let Y ∂S. Then we can find a monic⊆ linear pencil L of the same size of Y which∈ separates∈S from Y in the sense that L(A) 0 for all A S and ker L(Y ) = 0 . ≻ ∈ 6 { } There are two ways to prove this result. First, one can interpret matrix convex combinations as images of completely positive maps, an object originating from the theory of C∗-algebras. Hence one is able to use the power of this better developed area of mathematics to deal with matrix convex sets. The second option is to apply the Effros-Winkler separation technique which aims to translate separating linear forms into separating pencils. In this paper we present both ways. Material on completely positive maps, its connec- tions to matrix convexity and one proof of Lemma 2.1 are located in the appendix. We will need some of those results also later for the study of absolute extreme points. The Effros-Winkler separation technique is explained in the rest of this chapter starting after Remark 2.7. We will need some variations of this concept also later for the study of matrix exposed points. It is possible to understand most of the paper having read only one of those chapters. For readers unfamiliar with completely positive maps we recommend to read the chapter about the Effros-Winkler separation technique and to skim over the chapter in the appendix quickly. For readers familiar with completely positive maps we recommend to read this chapter until Remark 2.7 and the chapter in the appendix while only skimming over the chapter of the Effros-Winkler separation technique. We can reformulate the separation theorem by defining the notion of a polar of a matrix convex set and prove a bipolar theorem. Definition 2.2. Let S Sg be a matrix convex set. Then we define the free polar of S to be ⊆ g ◦ g g S = H S A S : I I Hi Ai 0 = H S S H . ∈ ∀ ∈ ⊗ − ⊗ { ∈ | ⊆ D } ( i ) X=1 ◦ ◦ S is a matrix convex set because S = L. L∈S D Theorem 2.3. [HKM1, Proposition 4.3]T (Bipolar theorem) Let S Sg. Then S◦◦ = ⊆ mconv(S 0 ). ∪ { } Proposition 2.4. [HKM1, Proposition 4.3] Let S Sg be a matrix convex set. Then 0 int(S◦) S is bounded. ⊆ ∈ ⇐⇒ Proof. [HKM1, Proposition 4.3] ” = ”: Let 0 int(S◦). Hence there is r > 0 such that ◦ ⇐ ∈ ◦ g B(0, r) S (1). In particular for i 1, ..., g we have rei S . Let A S . Then ⊆ ∈ { 2 2 } g± 2∈ g ∈ (1 reiX)(A)= I rAi 0. Therefore r Ai I and i=1 Ai r2 ± ± 11 P ” = ”: Let S be bounded. Fix r [0, ) such that A 2 r2 for all A S. Now let ⇒g 1 ∈ ∞ 2 ||2 2|| ≤1 ∈ 1 L S with L . We have that (Li Ai) L r 2 and hence Li Ai ∈ || || ≤ gr ⊗ || || g || ⊗ || ≤ g for all i 1, ..., g . Thus L(A) 0. ∈ { } Remark 2.5. Let S Sg be matrix convex and closed with 0 S. Then with the help of the bipolar theorem we⊆ get [0 int(S) S◦ is bounded] as a corollary∈ of Proposition 2.4. = can be proved without usage∈ of the⇐⇒ bipolar theorem by basically copying the second part⇒ of the proof of Proposition 2.4; we need this fact for the proof of Theorem 2.3 in the appendix. Proposition 2.6. Let S Sg be a matrix convex set. Then S◦(δ) = mconv(S(δ))◦(δ) ⊆ Proof. Clearly we have S◦(δ) mconv(S(δ))◦(δ). Now let H mconv(S(δ))◦(δ) and ⊆ ∈ assume there is some A S(k) and v Cδk such that v∗H(A)v < 0. If we write v = δ ∈ k ∈ k α=1 eα vα with vα C and define P : C im(P ) to be the projection onto ⊗ ∈ ∗ ∗ → ∗ M(A, v)H = span(v1, ..., vδ), then v H(P AP )v < 0, however P AP mconv(S(δ)). P ∈ Remark 2.7. In case that 0 / S in Lemma 2.1, the statements are still true if one deletes the word "monic". The reason∈ is that we can shift S and Y in such a way that 0 S. ∈ 2.1. A closer look at the Effros-Winkler separation technique. In this section we want to present and refine the separation technique of Effros and Winkler [WW]. The aim is to give a certificate that a point A Sg(δ) is not contained in a closed matrix convex set S by the means of a linear matrix inequality.∈ The idea is the following: One applies the usual Hahn-Banach separation to separate A from S(δ) with the help of an affine-linear function ℓ. Now one wants to translate this to a linear matrix inequality. By compressing elements of S to elements of S(δ) one is able to apply ℓ. Up to the constant part of ℓ this procedure translates ℓ into a homogeneous linear pencil LX with L Sg(δ). However translating the constant part is difficult and non-constructive. ∈ Therefore, to obtain the Effros-Winkler result for matrix cones is easier (see [FNT, Lemma 2.1] and Definition 6.4). So in order to separate A from S, it is convenient to homogenize the problem, separate the resulting matrix cone S′ from the point (I, A) and after dehomogenize (cf. Proposition 6.5). However it is not clear how one can homogenize ℓ. Therefore this approach has an even more non-constructive component. For our later results we have to strengthen the Effros-Winkler separation to be able to cover also situations where the matrix convex set is not closed (strong separation for non-closed sets) or only weak separation is possible (for instance separating a non-exposed extreme point x of a convex set S from S x ). The strong separation theorem for non-closed sets will be later needed to analyze matrix\ { } exposed points (cf. Theorem 6.20). Since the proof of the weak separation theorem is a bit technical and a slightly weaker statement can be recovered by using a homogenization trick, we have moved it to the appendix. We will use the weak separation to give a proof of the free Minkowski theorem in the appendix.

δ×δ Definition 2.8. We set δ = T SC T 0, tr(T ) = 1 . T { ∈ | } Proposition 2.9. [HM, Lemma 6.2] Let S Sg be matrix convex, 0 S and A Sg(δ) S. ⊆ ∈ ∈ \ Suppose there is a linear ϕ : Sg(δ) R such that ϕ(A) > 1 and ϕ(B) 1 for all → k×δ δ×δ ≤ B S(δ). For B S(k) and a contraction V C define fB,V : SC R, T ∈ ∗ ∗ ∈ ∈ k×δ → ∗ 7→ tr(V TV ) ϕ(V BV ). Then the set = fB,V k N, B S(k),V C ,V V I − F { | ∈ ∈ ∈ } is convex. For fB,V there is T δ such fB,V (T ) 0. ∈ F ∈T 12 ≥ δ−1 Proof. [HM, Lemma 6.2] Let fB,V . Choose w such that V w = V and set T = w w∗. Then T is positive∈F semidefinite and∈ Str(T ) = tr(w∗w||) =|| 1.|| We|| have ∗ ∗ ∗ ∗ ∗ 2 ∗ fB,V (T )= tr(V ww V ) ϕ(V BV )= tr((V w) V w) ϕ(V BV )= V ϕ(V BV )= ∗ − − || || − V 2 1 ϕ V B V 0. || || − ||V || ||V || ≥ k ×δ Let r N and Bj S(kj), Vj C j contractions, λj [0, 1] for j 1, ..., r such that r ∈ ∈ ∈ ∈ ∈ { } j=1 λj = 1. Define r P ∗ ∗ ∗ B = Bj and V = √λ1V1 . . . √λrVr j=1 M ∗ r ∗ ∗ r ∗ δ×δ Then B S, V BV = j=1 λjVj BjVj and V V = j=1 λjVj Vj I and for T SC ∈ ∗ ∗ r ∗ r ∗ ∈ we have tr(V TV ) = tr(TV V ) = j=1 λjtr(TVj Vj) = j=1 λjtr(VjTVj ). This shows r P P that fB,V = λjfB ,V and that is convex. j=1 j j PF P Lemma 2.10.P [HM, Lemma 6.1] Suppose is a convex set of affine-linear mappings F f : SCδ×δ R and that there is a concave function Ψ : R such that for all f → F → ∈ F there is T δ such that f(T ) Ψ(f). Then there is T δ such that f(T ) Ψ(f) for all f . ∈T ≥ ∈T ≥ ∈ F Proof. [HM, Lemma 6.1] δ is compact. Thus for f the set T δ f(T ) Ψ(f) T ∈ F { ∈ T | ≥ } is also compact. In order to show T δ f(T ) Ψ(f) = , it is enough to show f∈F { ∈T | ≥ } 6 ∅ that every finite intersection of those sets is non-empty. T m So let m N, f , ..., fm . We have to show that T δ fj(T ) Ψ(fj) = . ∈ 1 ∈ F j=1{ ∈ T | ≥ } 6 ∅ We set T m F : δ R : T (f (T ), ..., fm(T )). T → 7→ 1 m We want to show F ( δ) j=1[Ψ(fj), ) = . Assume the opposite is true. As F ( δ) is T ∩ ∞ 6 ∅ m T convex and compact, we can choose an affine linear h = j=1 hjXj + h0 and t ( , 0) m Q ∈ −∞ such that h( j=1[Ψ(fj), )) [0, ) and h(F ( δ)) ( ,t]. For each λ > 0 and ∞ ⊆ ∞ T ⊆P −∞ m k 1, ..., m we have 0 h((Ψ(f1), ..., Ψ(fm)) + λek)= h0 + j=1 hjΨ(fj)+ λhk which ∈ { }Q ≤ m implies hk 0. Without loss of generality we can suppose that j=1 hj = 1. Now set m ≥ P f = hjfj . j=1 ∈ F P m We knowP 0 h(Ψ(f1), ..., Ψ(fm)) = j=1 hjΨ(fj)+ h0. For every T δ we calculate m≤ m ∈ Tm f(T ) = hjfj(T ) = h(F (T )) h h(F (T )) + hjΨ(fj) < hjΨ(fj) j=1 − P0 ≤ j=1 j=1 ≤ Ψ(f), a contradiction. P P P Corollary 2.11. (Effros-Winkler - strong separation) [HM, Proposition 6.4] Let S Sg be matrix convex, 0 S and A Sg(δ) S. Suppose there is a linear ⊆ ∈ ∈ \ ϕ : Sg(δ) R such that ϕ(A) > 1 and ϕ(B) 1 for all B S(δ). Then there exists → ≤ ∈ T SCδ×δ and H Sg(δ) such that: ∈ ∈ For all B S : [T HX](B) 0, [T HX](A) 0 ∈ − − Proof. [HM, Proposition 6.4] We apply Proposition 2.9 and Lemma 2.10 with defined F like in Proposition 2.9 and Ψ := 0. We conclude that there is T δ such that f(T ) 0 for ∈T ≥ all f . Extend ϕ to a C-linear functional ϕ : (Cδ×δ)g C. By the Riesz representation ∈ F δ×δ → g ∗ theorem we can find matrices H , ..., Hg C such that ϕ(C) = tr(Hi Ci). It is 1 ∈ i=1 easy to see that the H have to be Hermitian. Hence H is Hermitian and HT = H . i i P i i g kδ δ k Now let B S (k) and v C . Write v = α=1 eα vα with vα C . Define the matrix ∈ ∈ 13 ⊗ ∈ P k×δ V = v1 . . . vδ C . We calculate (and denote the indices of a matrix as upper case letters) ∈ g ∗ ∗ ∗ v [T HX](B)v = v (T I)v v (Hi Bi)v − ⊗ − ⊗ i X=1 δ g δ = eα, T eβ vα, Ivβ eα,Hieβ vα,Bivβ h ih i− h ih i i α=1X,β=1 X=1 α=1X,β=1 δ g δ α,β ∗ α,β α,β ∗ α,β = T (V V ) H (V BiV ) − i i α=1X,β=1 X=1 α=1X,β=1 g g T ∗ T ∗ ∗ ∗ = tr(T V V ) tr(H V BiV ) = tr(V TV ) tr(HiV BiV )= fB,V (T ) − i − i i X=1 X=1 kδ−1 δ−1 δ If v and w we have V w α=1 wα vα 1 by the Cauchy- Schwarz∈ S inequality. This∈ S means that V ||is a contraction.|| ≤ | So| in || case|| ≤ that B S we have ∗ ∗ P δ ∈ v [T HX](B)v 0 because of V BV S. On the other hand let e = α=1 eα eα and − ≥ ∗ ∈ ⊗ set V = I (i.e. vα = eα). Then e [T HX](A)e = tr(T ) ϕ(A) < 0. − − P Proof. (of Lemma 2.1) (a) By the Hahn-Banach separation theorem from the theory of ordinary convexity we can find ϕ : Sg(δ) R linear such that ϕ(Y ) > 1 and ϕ(S(δ)) → ⊆ ( , 1]. Now we apply Corollary 2.11 to construct a pencil T HX such that for all −∞ − B S we have [T HX](B) 0 and [T HX](Y ) 0. There is ε > 0 such that ∈ − − [εI + T HX](Y ) 0. Now let D = √εI + T . Then I D−1HD−1X is the desired pencil. − − Proof. (of Theorem 2.3) It is easy to see that the polar is always matrix convex, closed and contains 0. Due to S S◦◦ this shows one direction. On the other hand suppose ⊆ A Sg(δ) such that A / mconv(S 0 ) =: T . T is also matrix convex and the Effros- ∈ ∈ ∪ { } g Winkler theorem says that there is L S (δ) such that T L however A / L. Hence L S◦ and A / S◦◦. ∈ ⊆ D ∈ D ∈ ∈ The next corollary is very important for the study of free spectrahedra. Corollary 2.12. [HKM1, Theorem 4.6, Proposition 4.9] Let L Sg. Then we have ◦ g ◦ ∈ L = H S L H = mconv(L, 0) and mconv(L) = L. If L is even bounded, Dthen mconv({ ∈ L, 0)| D = mconv(⊆ D } L). D D

Proof. First claim: Let H mconv(L, 0). This means there are matrices V1, ..., Vm such m ∗ ∈m ∗ m ∗ that j=1 Vj Vj I and j=1 Vj LVj = H. Let A L. Then H(A) = (I j=1 Vj Vj) m ∗ ∈ D ◦ − ⊗ I + (V I)L(A)(Vj I) 0. Hence mconv(L, 0) . Pj=1 j ⊗ P⊗ ⊆ DL P If AP Sg(δ) mconv(L, 0), we know from Lemma 2.1 that there exists H Sg(δ) such ∈ \ ∈ that 0 H(B) (I BX)(H) for all B mconv(L, 0) and 0 H(A) (I AX)(H). ≈ − ◦ ∈ ◦ ≈ − Thus we have H L and A / . This shows mconv(L, 0) . ∈ D ∈ DL ⊇ DL ◦◦ ◦ ◦ Second claim: By bipolarity we gain L = = mconv(L, 0) = mconv(L) . D DL Third claim: Now suppose L is bounded. We show that 0 mconv(L). We present the proof of [HKM5, PropositionD 4.2]. Let δ = size(L). In order∈ to show 0 mconv(L) set S = conv( v∗Lv v δ−1 ). We have to verify 0 S. Assume 0 / S. Then∈ there exists { | ∈ Sg } ∈ ∈ a linear functional ϕ : R R such that ϕ(S) R≤0. Set xi = ϕ(ei) for i 1, ..., g . Let → 14 ⊆ ∈ { } δ−1 ∗ g ∗ ∗ v , r > 0. Then we have v L(rx)v = 1 r ( xi(v Lv)i) = 1 rϕ(v Lv) > 0, ∈ S − i=1 − which contradicts the boundedness of L. D P We have seen how to separate a point from a closed matrix convex set with the help of a linear pencil. An exposed extreme point of a closed ordinary convex set S Rn is a point A S which can be separated from S A strictly with an aﬃne-linear⊆ function ψ (meaning∈ ψ(A) > ψ(B) for all B S A\). { However} the set S A is only closed if S = A . To characterize matrix exposed∈ \ { points} of matrix convex\ set { s,} we need some separation{ } theorem for non-closed matrix convex sets.

Definition 2.13. Let R be a real closed extension field of R and let a, b R> . We write ∈ 0 a >> b if a > Nb for all N N. For b R with an N N such that N < b < N we ∈ ∈ ∈ − denote by st(b) R the standard part of b (a concrete definition is given in the appendix). ∈

∗ i On the algebraic closure C = R[i] we have an involution : C C, a + b a r×k →∗ 7→ − bi (a, b R), which leaves R invariant. For A C we obtain A by transposing A and afterwards∈ applying the involution to each entry.∈ If A = A∗, we call A Hermitian and define SCk×k = A Ck×k A∗ = A . We remark that this notion depends not only on C but also the choice{ ∈ of the| real closed} field R, however this should not cause ambiguity δ×δ ∗ anywhere. We set R,δ = T SC T 0, tr(T ) = 1 where T 0 means v T v 0 T { ∈ | } ≥ for all v Cδ. For A SCδ×δ we write A 0 if v∗Av 0 for all v Cδ and A 0 if ∈ ∈ R ≥ ∈ ≻R v∗Av > 0 for all v Cδ 0 . ∈ \ { } Lemma 2.14. Let be a real closed extension field of R and v g. Then there exist R j r j ∈ R r 1, ..., g , λ , ..., λr R and a R such that vi = λja for all i 1, ..., g and ∈ { } 1 ∈ i ∈ j=1 i ∈ { } λ1 >> ... >> λr > 0. P Proof. Without loss of generality v1 > ... > vg > 0. Set λ1 = v1. Then write (0, w) = vg v λ1(1, ...., st ). For every entry wj of w we have λ1 >> wj. Now continue inductively − v1 with w. Proposition 2.15. Let S Sg be matrix convex, 0 S and A Sg(δ) S. Suppose there ⊆ ∈ ∈ \ is a linear ϕ : Sg(δ) R such that ϕ(A) = 1 and ϕ(B) < 1 for all B S(δ). For B S(k) → k×δ δ×δ ∈∗ ∗ ∈ and a contraction V C define fB,V : SC R, T tr(V TV ) ϕ(V BV ). Then ∈ k×δ →∗ 7→ − the set = fB,V k N, B S(k),V C ,V V I is convex. For fB,V there F { | ∈ ∈ ∈ } ∈ F is T δ such fB,V (T ) > 0. ∈T Proof. This is basically the same proof as the one of Proposition 2.9. Lemma 2.16. Suppose is a convex set of affine-linear mappings f : SCδ×δ R such F → that for all f there is T δ such that f(T ) > 0. Then there exists a real closed field ∈ F ∈T extension R of R and T R,δ such that f(T ) > 0 for all f (f has a unique extension ∈T δ×δ ∈ F to an R-affine linear map SR[i] R). → Proof. For f consider the R-semialgebraic classes ∈ F (R, T ) R real closed extension field of R, T R,δ,f(T ) > 0 . { | ∈T } Theorem X.29 tells us that in order to prove

(R, T ) R real closed extension field of R, T R,δ,f(T ) > 0 = { | ∈T } 6 ∅ f\∈F it is enough to show that every finite intersection of those sets is non-empty. That the latter is the case is basically the same proof as in Lemma 2.10. 15 Corollary 2.17. (Effros-Winkler separation for non-closed sets) Let S Sg be matrix ⊆ convex, 0 S and A Sg(δ) S. Suppose there is a linear ϕ : Sg(δ) R such that ∈ ∈ \ → ϕ(A) = 1 and ϕ(B) < 1 for all B S(δ). Then there exists T SCδ×δ and H Sg(δ) such that: ∈ ∈ ∈ For all B S : [T HX](B) 0, ker[T HX](A) = 0 ∈ − ≻ − 6 { } Proof. The same proof as the one of Corollary 2.11 gives us a real closed extension field δ×δ g R of R with algebraic closure C = R[i], T SC and H S (δ) such that T 0, tr(T ) = 1 and ∈ ∈ for all B S : [T HX](B) 0, e∗[T HX](A)e = 0, ∈ − ≻R − δ δ×δ 2 where e = α=1 eα eα. Since T 0, we can find D SC such that T = D . ⊗ ∈ δ×δ With Lemma 2.14 we find r N and λ1 >> ... >> λr > 0 R, Dj SC such that r P ∈ ∈ ∈ D = λjDj. In case that r = 1, we are done. So suppose that r 2. As tr(D) = 1, j=1 ≥ it is easy to see that we can choose λ = 1. Set E = r−1 λ D . We want to replace P 1 j=1 j j T = D2 by E2 + D2. r P 2 Indeed let B S. Applying the standard part, we see that [D1 HX](B) R 0. Now let δ2 ∈ ∗ 2 2 − ∗ 2 v C 0 and assume v [E + Dr HX](B)v 0. We know v [D1 HX](B)v 0 ∈ ∗ \2 { } − ≤ ∗ 2 2 − ≥ and v [Dr ](B)v 0. The sum of those two terms is st(v [E + Dr HX](B)v) 0, thus ∗ 2 ≥∗ 2 − ≤ v (D I)v = v [D ](B)v = 0. However this means that (Dr I)v = 0. We conclude r ⊗ r ⊗ that v∗[E2 + D2 HX](B)v = v∗[T HX](B)v > 0, a contradiction. We have shown r − − [E2 + D2 HX](B) 0 for all B S. r − ≻R ∈ ∗ 2 On the other hand we have e [T HX](A)e = 0. Clearly Rλr spanR( λjλh j, h − ∩∗ 2 { | ∈ 1, ..., r 1 λjλr j 1, ..., r 1 )= 0 which means 0= e [Dr ](A)e = 0. Again { − }} ∪ { | ∈ { ∗ −2 }} 2 { } ∗ we infer that (Dr I)e = 0 and e [E + Dr HX](A)e = e [T HX](A)e = 0. We have ⊗ 2 2 − − shown that we can replace T by E + Dr . In the same way we continue and show that we 2 2 r−2 2 2 2 can replace E + Dr by ( j=1 λjDj) + Dr + Dr−1 and so on. Inductively we deduce that r D2 HX is the desired pencil. j=1 j − P PProof. (of Lemma 2.1 (b)) From the theory of convexity we know that there exists ϕ : Sg(δ) R linear such that ϕ(B) < 1 for all B S(δ) and ϕ(Y ) = 1. Now Corollary 2.17 → ∈ tells us that there exists T SCδ×δ and H Sg(δ) such that: ∈ ∈ For all B S : [T HX](B) 0, ker([T HX](A)) = 0 ∈ − ≻ − 6 { } Proposition 1.12 tells us that we can replace T HX by a monic linear pencil L. − 3. Projection number and convexity number The following definition is a generalization of the concept of the minimal and maximal matrix convex set generated by a subset of Rg, which was covered in [DDSS, Chapter 4] and in [FNT]. The corresponding results up to Lemma 3.6 are present in [DDSS]. g Definition 3.1. Let T = (Tn)n S be closed. Define for m N the sets [Example 7.6] ∈N ⊆ ∈ pz (T ) := ( B Sg(k) m { ∈ | For all projections P : Ck im P of rank m : P BP ∗ T (rk(P )) → ≤ ∈ k∈N kzm(T ) := mconv (T (1), ..., T (m)) o 16 For matrix convex T we set pz(T ) := inf m N pz (T )= T N the projection number of T { ∈ | m }∈ ∪ {∞} kz(T ) := inf m N kzm(T )= T N the convexity number of T { ∈ | }∈ ∪ {∞} g Proposition 3.2. Let T S be a closed matrix convex set. Then kzm(T ) is the smallest matrix convex set S with⊆S(j)= T (j) for j 1, ..., m . On the other hand, S = pz (T ) ∈ { } m is the largest set satisfying this. If 0 T , then pz (T )= g L. ∈ m L∈S (m),T (m)⊆DL D Proof. The claim concerning kzm(T ) is true by definition.T For the other we suppose WLOG that 0 T (otherwise consider a shifted version of T ; we remind the reader that T (1) is non-empty∈ because T is closed with respect to compressions). We show that pz (T ) = g L. Let A pz (T ). If we m L∈S (m),T (m)⊆DL D ∈ m had L Sg(m) with L(A) 0, then Lemma 1.13 would give us a projection P to an at ∈ T most m-dimensional subspace with L(P AP ∗) 0 as well. Since P AP ∗ T (rk(P )), we ∈ would conclude T (rk(P )) * L. D Conversely let A / pz (T ) and take a projection P with P AP ∗ / T . By the Effros- ∈ m ∈ Winkler separation method Lemma 2.1 we find L Sg(m) such that L separates P AP ∗ ∗ ∈ ∗ from T (i.e. T L and P AP / L). We can interpret P AP a submatrix of A (up ⊆ D ∈ D to unitary equivalence). Since L(P AP ∗) is a submatrix of L(A) and L(P AP ∗) 0, also L(A) 0.

This shows that pzm(T ) is matrix convex as an intersection of matrix convex sets. From the Eﬀros-Winkler separation it is clear that that it contains T (j)=pz (T )(j) for j m. m ≤ That it is the largest possible set is due to the fact that for all A Sg(k) and every ∈ projection P : Ck im(P ) we have P AP ∗ mconv(A). → ∈ g Corollary 3.3. (cf. [HM, Lemma 7.3]) Let L S (k). Then pz( L) k. ∈ D ≤ Proof. This is an immediate corollary of Lemma 1.13. Proposition 3.4. Let T Sg, m N and suppose that for all B T , k N and ⊆ ∈ ∈ ∈ projections P : Ck im(P ) of rank at most m we have P BP ∗ T . Then mconv(T )(m)= mconv(T (m))(m). → ∈ k ×m r ∗ Proof. Let A mconv(T )(m) and Vj C j , Bj T (kj) with V Vj = I and ∈ ∈ ∈ j=1 j r V ∗B V = A. Let P be the projection of Ckj onto im(V ). Then we have j=1 j j j j j P r r P ∗ ∗ ∗ ∗ A = V BjVj = V P (Pj BjP )PjVj mconv(T (m)). j j j j ∈ j j X=1 X=1 Corollary 3.5. Let T Sg be a closed matrix convex set. Then we have ⊆ pz (T )= A Sg (mconv(A))(m) T m { ∈ | ⊆ } Proof. This follows from Proposition 3.4. g ◦ Lemma 3.6. Let T S be closed and matrix convex with 0 T . Then kzm(T ) = ◦ ◦⊆ ◦ ∈ ◦ ◦ pz (T ) and pz (T ) = kzm(T ). If 0 int(T ) we also have pz (T ) = kzm(T ). m m ∈ m Proof.

◦ (a) pzm(T ) = L = L = L  g ◦ D   ◦◦ D   D  L∈S (m\),T ⊆DL L∈T\(m) L∈\T (m)   17     ◦ (b) ◦ ◦ = mconv(L) = mconv(L) = kzm(T )     L∈\T (m) L∈[T (m)     where the equalities follow from these principles: (a) We know that

◦ L = L = pzm(T ) g ◦ D g ◦ D L∈S (m\),T ⊆DL L∈S (m)\,T (m)⊆DL where we used Lemma 1.13 and the description obtained in Proposition 3.2. (b) This is Corollary 2.12. ◦ ◦◦ ◦ ◦ ◦◦ ◦ For the other claim we calculate pzm(T ) = pzm(T ) = kzm(T ) = kzm(T ). In case that 0 int(T ) we know from Proposition 2.4 that T ◦(m) is compact. Thus the theorem ∈ ◦ of Caratheodory Lemma 1.8 implies that kzm(T ) is closed. Corollary 3.7. Let T Sg be matrix convex and closed with 0 T . Then kz(T ) pz(T ◦) and pz(T ) kz(T ◦). In⊆ case that 0 int(T ), we have pz(T ) =∈ kz(T ◦). In case≥ that T is bounded, we≤ have kz(T ) = pz(T ◦). ∈

g Proof. Let S S be matrix convex and closed with 0 S. Suppose that kzm(S) = S. ⊆ ◦ ◦ ◦ ∈ ◦ ◦ Then we know S = kzm(S) = pzm(S ). This shows kz(T ) pz(T ) and pz(T ) kz(T ). ◦ ≥ ◦ ≤◦ Now in case that 0 int S and pzm(S) = S, then S = pzm(S) = kzm(S ). Asa consequence we get kz(∈ T ◦) pz(T ) in case that 0 int T . ≤ ∈ In case that T is bounded, we know that 0 int T ◦ and therefore kz(T ) = kz(T ◦◦) pz(T ◦). ∈ ≤

The next result characterizes how small the maximal size of the blocks occuring in the block diagonalization of a pencil description of a given free spectrahedron can be. We encourage the reader to compare this result with [FNT, Theorem 2.3].

Proposition 3.8. Let T = L be a free spectrahedron given by a monic linear pencil L. Then pz(T ) k if and onlyD if T can be written as a finite intersection of spectrahedra defined by monic≤ pencils of size k. Proof. " =": Suppose T is a finite intersection of spectrahedra defined by monic linear pencils of⇐ size k. Then Corollary 3.3 tells us that each of these spectrahedra has projection number at most k. This means that pz( L) k. D ≤ "= ": Let m := pz(T ) k. Since 0 int T , Corollary 3.7 tells us kz(mconv(L, 0)) = ⇒◦ ≤ ∈ kz(T ) = pz(T ) = m k. Therefore there exist H , ..., Hr mconv(L, 0)(k) such that ≤ 1 ∈ L 0 mconv(H1, ..., Hr). This means mconv(H1, ..., Hr) = mconv(L, 0). We conclude ⊕ ∈ r r T = I H X = j=1 Hj (Corollary 2.12). D −Lj=1 j D g Corollary 3.9. Let T = L S be a free spectrahedron given by a monic linear pencil L. Then pz(T ) is the leastD number⊆ k for which we find finitely many spectrahedra each defined by a monic pencil of size at most k whose intersection is T . The following corollary was already proven in [FNT, Theorem 3.2], however that proof did not use polarity.

Corollary 3.10. [FNT, Theorem 3.2] Let S Rg = Sg(1) be convex with 0 int(S). ⊆ ∈ Then pz1(S) is a free spectrahedron if and only if S is a polyhedron. 18 Proof. Let pz1(S) be a spectrahedron. Corollary 3.9 yields that pz1(S) is a intersection of spectrahedra defined by 1 1 matrices. So S is a polyhedron. × g If S is a polyhedron, then we find f , ..., fr R[X] with S = x R 1+ f (x) 1 ∈ 1 { ∈ | 1 ≥ 0, ..., 1+ fr 0 . The free spectrahedron defined by the monic pencil L with the 1+ fj ≥ } on the diagonal satisfies L(1) = S. Of course we also have pz( L) = 1 (Corollary 3.3); D D thus L = pz (S). D 1 Remark 3.11. (cf. [DDSS, Proposition 3.5]) Let T Sg be matrix convex, compact ⊆ and kz(T ) = m < . Then there exists a dense subset Ln n N of T (m) such ∞ { | ∈ } that mconv( n∈N Ln) = T (cf. Lemma X.23 for more details). By taking the polar (Remark 2.5 and Corollary X.22) we get the following statement: L Let S Sg be matrix convex, closed and with 0 int S as well as pz(S)= m< . Then ⊆ ◦ ∈ ∞ there exists a dense subset Ln n N of S (m) such that L = S. { | ∈ } DLn∈N n g Proposition 3.12. Let L S (δ) such that L is bounded. Let be the operator system [Example 7.1] ∈ D T defined by the Li and let m N. Then kz( L) m if and only if: ∈ D ≤ For all finite-dimensional C∗-algebras and all unital m-positive maps ϕ : the map ϕ is already completely positive. B T →B ∗ Proof. "= ": Let kz( L) m. Fix a finite-dimensional C -algebra and suppose that ⇒ D ≤ B ϕ : is m-positive. Let H = I ϕ(L )X ... ϕ(Lg)Xg. Theorem X.19 tells us T →B − 1 1 − − that L(m) H(m). Since kz( L) m, we conclude L H. Again Theorem X.19 tells usD that ⊆ϕ is D completely positive.D ≤ D ⊆ D

" =": Assume the right hand side holds and that kz( L) >m. Since kzm( L) is closed, the⇐ Effros-Winkler separation Lemma 2.1 (a) tells us thatD we can find a monicD linear pencil H, s N and B L(m + s) such that L(m) H(m) and B / H. Let be the ∈ ∈ D D ⊆ D ∈ D R operator system defined by the Hi. Theorem X.19 implies that the linear map ϕ : T → R given by Li Hi and I I is m-positive. By assumption it is also completely positive 7→ 7→ so B L H. ∈ D ⊆ D 4. Helton-McCulloughs characterisation of free spectrahedra 4.1. Infinite-dimensional projection lemma and Nash manifolds. Definition 4.1. Let p S(C X )δ×δ be a Hermitian matrix polynomial. For B Sg(r) ∈ h i ∈ with p(B) 0 we define for all k N the set p(B)(k) as the connected component of ≻ kr×kr g ∈ E p(kr) := A (SC ) p(A) is invertible containing Ik B (we have p(Ik B) 0 I { ∈ | } ⊗ ⊗ ≻ automatically, see Remark 4.6). Denote its closure by p(B)(k). If p is a monic linear D pencil, then p(0) = p (Proposition 4.3). D D We want to prove the following: δ×δ Theorem 4.2. [HM, Theorem 1.4] Let δ N, p SC X of degree d, S := p(0). [Example 7.2] ∈ ∈ h i D Suppose that p(0) = Iδ and that p(0) is matrix convex. Then S is a free spectrahedron. E Furthermore S is a finite intersection of spectrahedra defined by linear matrix inequalities 1×δ of size k := dim C X d . h i 2 ⌊ ⌋ This result was originally proven by Helton and McCullough and is formulated for bounded S although that assumption is not used in their proof. In [HM] a theory of varieties on ∂S was introduced. Building on that, they constructed a monic linear pencil representing S by applying the Effros-Winkler separation finitely many times. In each step one separates one 19 point X ∂S from S with a monic linear pencil L such that int S int L. By using the Noetherian∈ character of their variety-constructions, Helton and McCullough⊆ D showed that if the point X is chosen well, then all other points which have the same "vanishing ideals" get also separated and one needs only finitely many steps to separate ∂S from int S. Our strategy is to merge infinitely many points in ∂S to an operator point (see Lemma 4.7 for the details) and separate this from int S with a monic linear pencil L. If the matrix points belong to the same cell of a cell-decomposition of ∂S(k) (more precisely the cell-decomposition of a slightly more complicated set), then L separates all these points simultaneously. While modifying the main step of Heltons and McCulloughs argument significantly, we adopt the general framework of their proof; thus one should regard our proof as a new variant of theirs rather than to be completely different. The size of the LMI needed in Theorem 4.2 can be estimated by the product of k and the numbers of cells in our cell-decomposition. However it is hard to estimate the latter number. The proof in [HM] gives a cleaner estimate for the size of the LMI. δ×δ Proposition 4.3. [HM, Lemma 2.1] Let p SC X such that p(0) 0 and p(0) is matrix convex. Then ∈ h i ≻ E g ∂ p(0) = p(0) p(0) = A S ker p(A) = 0 , t (0, 1) : tA p(0) . D D \E { ∈ | 6 { } ∀ ∈ ∈E } Proof. (cf. [HM, Lemma 2.1]) Let A p(0) p(0). By continuity we have p(A) 0. By ∈ D \E matrix convexity of p(0) and continuity, we even get that p(tA) 0 and tA p(0) for all E ∈ D t (0, 1). The polynomial function ψ : R R,t det(p(tA)) fulfills ψ([0, 1]) R≥0 and ψ∈(0) > 0. Hence there are only finitely many→ t 7→(0, 1) such that ψ(t) = 0. Let⊆s (0, 1) ∈ ∈ and fix t (s, 1) with ψ(t) = 0. Then tA lies in p(0) p(0). However p(0) is closed in ∈ 6 I ∩ E E p(0) by definition, thus tA p(0) and by convexity also sA p(0). Now we see that I ∈ E ∈ E A / p(0) implies that find a non-trivial vector v ker p(A). ∈E ∈ g Corollary 4.4. Let L be a monic linear pencil. Then int L = A S L(A) 0 and D { ∈ | ≻ } int L = L. D D Definition 4.5. Let p SC X δ×δ be a matrix polynomial of degree d with p(0) = g ∈ kδ h i δ k Iδ. Let A S (k), v C with p(A)v = 0. Write v = eα vα with vα C ∈ ∈ α=1 ⊗ ∈ δ 1×δ and eα C the α-th unit vector. Then define M(A, v)p :=P q(A)v q C X d = ∈ ∈ h i⌊ 2 ⌋ span q(A)vα q C X ⌊ d ⌋, α 1, ..., δ . If d = 1 (so p is a linear pencil), notice ∈ h i 2 ∈ { } that then equality M(A, v)p = span v1, ..., vδ oholds. { }

Remark 4.6. Let , be Hilbert spaces, A ( ), (Bj)j ( ) and (eα)α A be a H K ∈B H ∈N ⊆B K ∈ j (∞) Hilbert space basis of and w = eα wα . Then we have H α∈A ⊗ j∈N ∈H⊗K P L ∗ A Bj = U A Bj U and ⊗    ⊗  Mj∈N Mj∈N     ∗ ∗ j Ai Bj w = U A Bj Uw = U (A Bj) eα wα  ⊗    ⊗  ⊗ ⊗ !! Mj∈N Mj∈N Mj∈N αX∈A where U is the unitary shuﬄe operator deﬁned by

(∞) (∞) j j U : ( ) , eα vα eα vα . H ⊗ K → H ⊗ K ⊗   7→ ⊗ ! αX∈A Mj∈N Mj∈N αX∈A 20   δ×δ We infer that for p C X and (An)n (K) we have p An p(An). ∈ h i ∈N ⊆B n∈N ≈ n∈N Lemma 4.7. (Inﬁnite-dimensional projection lemma) SupposeLp SC X Lδ×δ is a ma- g∈ h i trix polynomial of degree d with p(0) = Iδ. Let (Xn)n∈N S (h) be a bounded se- hδ−1 ⊆ δ α quence and (vn)n∈N such that p(Xn)vn = 0. Write vn = α=1 eα vn where δ ⊆ S h (∞) ⊗δ α eα C is the α-th unit vector. Set B = n∈N Xn ((C ) ) and v = α=1 eα v α P ∈ α vn h (∞) ∈ B ⊗ where v := 2 (C ) . Then if P denotes the projection onto M(B, v)p := n∈N n ∈ L P 1×δ ∗ q(B)v q LC X d , we have also v,p(P BP )v = 0. ∈ h i⌊ 2 ⌋ h i 1×δ ∗ Let k = dim C X d . If p(0) is matrix convex and all Xn p(0), then even p(P BP )v = h i⌊ 2 ⌋ E ∈ D k×k k×δ 0. In this case there exist monic L C X 1 , W C X d such that p(0) L and ∈ h i ∈ h i⌊ 2 ⌋ D ⊆ D L(B)W (B)v = 0 while W (B)v = 0. This means L(Xn)W (Xn)vn = 0 for all n N and 6 ∈ W (Xn)vn = 0 for one n N. (II) 6 ∈ Proof. This proof is based on [HM, Lemma 7.3]. Let m C X be a monomial of degree ∈d h i at most d. Write m = w1xjw2 with deg(w1), deg(w2) 2 . Now for α, β 1, ..., δ we ∗ α β ≤ ⌊ ⌋ ∈ { } have w (B)v , w (B)v M(B, v)p by construction. Thus 1 2 ∈ β ∗ α ∗ ∗ β α v ,m(P BP )v = P P w (B) v ,Bjw (B)v h i h 1 2 i ∗ β α β α = w (B) v ,Bjw (B)v = v ,m(B)v h 1 2 i h i

We write p = γ∈Fg fγγ and calculate P δ ∗ β ∗ α v,p(P BP )v = eβ v , (fγ γ(P BP ))(eα v ) h i h ⊗ ⊗ ⊗ i γX∈Fg α,βX=1 δ δ β ∗ α β α = eβ,fγ eα v , γ(P BP )v = eβ,fγ eα v , γ(B)v h ih i h ih i γX∈Fg α,βX=1 γX∈Fg α,βX=1 = v,p(B)v = 0. h i Now suppose that all Xn p(0) and that p(0) is matrix convex. For s N consider s∈ D E s→∞ ∈ ∗ s→∞ the partial sum Bs = n=1 Xn 0. Then we see that Bs ⇀ B and PBsP ⇀ ⊕ s P BP ∗ (where ⇀ denotes weak convergence). Thus also p(PB P ∗) →∞⇀ p(P BP ∗) and L s p(P BP ∗) 0 due to matrix convexity. We have already seen v,p(P BP ∗)v = 0. Both facts together mean that p(P BP ∗)v = 0. By the Eﬀros-Winklerh separationi Lemma 2.1 k×k ∗ (b) there exists monic L SC X 1 such that p(0) int L and ker L(P BP ) = 0 . ∈ h i E ⊆ k×δD 6 { } Choose a non-trivial kernel vector w and ﬁnd W C X d such that w = W (B)v. So ∈ h i⌊ 2 ⌋ we have 0= L(P BP ∗)W (B)v. This translates into 0 = W (B)v, L(P BP ∗)W (B)v = W (B)v, L(B)W (B)v . Together s→∞ h i h i with L(Bs) ⇀ L(B) and thus L(B) 0, this means L(B)W (B)v = 0. Since NC-matrix polynomials are behaving well with direct sums (see Remark 4.6), we get L(Xn)W (Xn)vn = 0 for all n N. ∈ Remark 4.8. The conclusion (II) of Lemma 4.7 remains true even in the case that the sequence (Xn)n∈N is not bounded.

Proof. By applying Lemma 4.7 for a finite number of Xn we get the following: For every k×δ k×k fixed number t N there exists W C X d , L C X 1 such that S int L and ∈ ∈ h i⌊ 2 ⌋ ∈ h i ⊆ D 21 L(Xn)W (Xn)vn = 0 (*) for all n 1, ..., t as well as W (Xn)vn = 0 for one n. ∈ { } 6 k×δ Now LW C X d lives in a finite-dimensional subspace. If we enlarge t, the subspace ∈ h i⌊ 2 ⌋+1 of possible solutions to ( ) is becoming smaller. Since we find a solution for arbitrary big ∗ t, there exists a common solution L, W for all n N. ∈ Remark 4.9. For our new proof of Theorem 4.2 we need the concept of a Nash submanifold of Rk. A Nash submanifold M Rk (of dimension r) is a connected submanifold of Rk for which for each x M there exists⊆ a semialgebraic C∞-diffeomorphism map φ : U H ∈ → of the atlas of M from a semialgebraic neighborhood U of x in Rk to a semialgebraic neighborhood H of 0 in Rk with ϕ(M U) = H (Rr 0 k−r). We refer the reader to [BCR] for an introduction to Nash manifolds.∩ ∩ A Nash× {function} is a function from a Nash manifold into R whose graph is a Nash manifold. A function f : Rn Rm is a Nash → function if and only if f C∞ and the graph of f is semialgebraic. ∈ We need two facts: The first is that each semialgebraic set admits a decomposition into finitely many Nash manifolds C∞-diffeomorphic to relatively open boxes [BCR, Proposition 2.9.10]. This is called analytic/C∞ cell decomposition. The second fact is that a Nash function f defined on an open box has locally a converging power series expansion [BCR, Chapter 8.1]. In particular the set f −1(R 0 ) is dense in the domain of f if f = 0. \ { } 6 In the following we will identify C with R2 and are thereby able to talk about semialgebraic subsets of C. Lemma 4.10. In the setting of Theorem 4.2 ∂S(k) decomposes into a finite disjoint union of Nash manifolds Nash-diffeomorphic to relatively open boxes. Additionally we can achieve kδ−1 that there is a a Nash function fC on every box which satisfies fC (A) ker(p(A)) ∈ S ∩ Proof. As ∂S(k) is a semialgebraic set over R (Proposition 4.3), we can make an analytic cell decomposition which gives the first statement. For the second part we choose a semialgebraic function fC with the desired property and apply the analytic cell decomposition theorem again. 4.2. Proof of Theorem 4.2. Proof. (of Theorem 4.2) We write ∂S(k) as in Lemma 4.10. Then we consider one Nash manifold C with associated Nash function fC . Fix a dense subset (Xn)n∈N of C. We apply k×k k×δ Lemma 4.7 which says that there exists L := LC SC X 1 monic and V C X d ∈ h i ∈ h i⌊ 2 ⌋ such that int S int L, L(Xn)V (Xn)fC (Xn) = 0 for all n N and V (Xm)fC (Xm) = 0 ⊆ D ∈ k2 6 for one m. Therefore we conclude that the function φ : C C , X V (X)fC (X) is → 7→ not zero and L(X)V (X)fC (X) = 0 for all X C. Since C is a Nash manifold and V is a matrix polynomial, we know that V is in each∈ component a Nash function on C. We can assume without loss of generality that C is a relatively open box. 2 2 But then Remark 4.9 implies that φ−1(Ck 0 ) is dense in Ck and therefore L(X) has non-trivial kernel for all X C. This means\ that{ } L separates all points of C from int(S). ∈ Therefore the direct sum of all LC separates all points of ∂S(k) from int(S). We have pz(S) k (indeed, let A ∂S(h) and v δh with p(A)v = 0. One can adapt the first ≤ ∈ ∈ S calculations of the proof of Lemma 4.7 to obtain P AP ∗ ∂S where P : Ch im(P ) ∈ → is the projection onto M(A, v)p which proves the claim. Alternatively we can apply Lemma 4.7 to the constant sequences B = (A)n∈N and (v)n∈N and obtain: There ex- k×k k×δ ist monic L C X 1 , W C X d such that S L and L(A)W (A)v = 0 while ∈ h i ∈ h i⌊ 2 ⌋ ⊆ D h W (A)v = 0. Thus A ∂ L(h) and there exists a projection P : C im(P ) of rank at 6 ∈ D 22 → ∗ ∗ most k such that P AP ∂ L (Corollary 3.3). Since P AP S and S L, we deduce P AP ∗ ∂S). ∈ D ∈ ⊆ D ∈ Since pz( L ) k (Corollary 3.3) for all C and pz(S) k and the interior of the spec- D C ≤ ≤ trahedron defined by the direct sum of the LC agrees with int S on the level k (because of Proposition 4.3 and Lemma 4.7), the claim follows.

In [HM, Remark 1.1] Helton and McCullough say that if we replace x = 0 in S = p(x) by D another point x Rn, then S is still a free spectrahedron. This is simply done by a change of coordinates. However∈ if we consider a matrix point x, they point towards an extension of the methods of their proof. We show in the following that the situation for an arbitrary matrix point can be reduced to the situation of a scalar point. The reduction to the scalar case involves to check some extra property, which seems to be not more difficult to check than the usual requirements of the scalar case. Corollary 4.11. Let p SC X δ×δ be a matrix polynomial, A Sg(n) and p(A) 0. ∈ h i ∈ ≻ Then the set p(A) constitutes the nN-th levels of the interior of a spectrahedron E DB−CX with (B CX)(A) 0 if and only if p(A ) 0, the line between A and In A is in p − ≻ 11 ≻ ⊗ 11 I and p(A ) is matrix convex. E 11 Proof. " =:" The right hand side implies that p(A ) equals a spectrahedron ⇐ D 11 DB−CX such that (B CX)(A ) 0. We have Ir A p(A )(nr) and therefore p(A)(r) = − 11 ≻ ⊗ ∈ E 11 E p(A )(nr). From Proposition 4.3 we know that int( p(A )) = p(A ). E 11 D 11 E 11 "= :" Suppose p(A)(r) = int( (nr)) for all r N for a pencil B CX with ⇒ E DB−CX ∈ − (B CX)(A) 0. Since int( ) is matrix convex and A = eT Ae mconv(A), we − ≻ DB−CX 11 1 1 ∈ conclude A int( ). Thus In A int (n)= p(A)(n). Since p behaves 11 ∈ DB−CX ⊗ 11 ⊆ DB−CX E well with direct sums, we get A p, p(A ) 0 and p(Inr A ) 0 for all r N. 11 ∈ I 11 ≻ ⊗ 11 ≻ ∈ int( (nr)) is convex so the line between Ir A and Inr A is in p(A). Now we DB−CX ⊗ ⊗ 11 I have p(A )(nr)= p(Inr A )(1) = int (nr) for all n N. int is matrix E 11 E ⊗ 11 DB−CX ∈ DB−CX convex and p(A11) is clearly closed with respect to direct sums and reducing subspaces (the latter oneE follows from the fact that submatrices of positive definite matrices are again positive definite). We conclude p(A11) = int( B−CX ) (if we calculate with matrices of size = nk, we can lift them to aE multiple of k byD making the direct sum with sufficiently 6 many A11).

So we know that p(A11) is the interior of a spectrahedron and in particular matrix convex. E

5. -irreducible pencils and the Gleichstellensatz D 5.1. Sequence of determinants of a monic linear pencil. In this subsection ﬁx a monic linear pencil L generating a spectrahedron L which has on each level the determi- 1 g D nant fk = detk L( ) R[ , ..., α, β 1, ..., k ]. X ∈ Xα,β Xα,β | ∈ { } Lemma 5.1. [HKV, Lemma 2.1] Let f R[ ℓ α, β 1, ..., k , 1 ℓ g] be in- ∈ Xα,β | ∈ { } ≤ ≤ variant under unitary conjugations. Then also all factors of f are invariant under unitary conjugations.

Proof. (cf. [HKV, Lemma 2.1]) Let f be monic and f = p ... pm C[ ] be the 1 · · ∈ X decomposition of f into irreducible polynomials over C in such a way that associated polynomials in the factorization are equal. Since the set of unitary matrices in Ck×k is connected, and the map C[ ] Ck×k C[ ], (p,Y ) (A p(Y ∗AY )) is continuous, X × → 23X 7→ 7→ k×k we conclude that for each unitary U C and h 1, ..., m there exists λh(U) C ∗ ∈ g ∈ { } ∈ such that ph(UAU ) = λh(U)ph(A) for all A S (k). We show that λ(U) := λ1(U) = 1 for every unitary U. ∈ Let diag : Ck Ck×k be the map that maps a vector v to the diagonal matrix V with v → k×k on the diagonal. The map λ is multiplicative. Since Ik is in the commutator of C we have λ(aI) = 1 for every a 0. For unitary U, V Ck×k we have λ(UV )= λ(U)λ(V )= λ(V )λ(U)= λ(VU). We conclude∈ S λ(U ∗VU)= λ(V∈); thus λ is determined by its restriction 0 0 to diagonal matrices and is constant on unitary orbits. For a1, ..., ak choose a k k ∈ S ∈ S such that a = j=1 aj. Then we have

Q k k λ(diag(a1, ..., ak)) = λ(diag(1, ..., 1, aj , 1, ..., 1)) = λ(diag(aj, 1, ..., 1)) j=1 j=1 Y j-th index Y k |{z} k =λ(diag(ak, 1, ..., 1)) = λ(diag(a, 1, ..., 1)) = λ(diag(1, ...1, a , 1, ..., 1)) j j Y=1 Y=1 j-th index =λ(diag(a, ..., a)) = 1 |{z}

Lemma 5.2. (Rule of Descartes for RZ-polynomials) Let (bj)j∈{1,...,d} be a strictly increas- d bj ing tuple of natural numbers and 0, f = ab X R[X] 0 be a real polynomial such j=1 j ∈ \{ } that a R 0 for all j. We call j 1, ..., d 1 a a < 0 the number of sign bj P bj bj+1 ∈ \ { } |{ ∈ { − } | }|bj +bj changes of the coefficient sequence of f and j 1, ..., d 1 ( 1) +1 ab ab < 0 |{ ∈ { − } | − j j+1 }| the number of sign changes of the coefficient sequence of f( X). For x R we denote − ∈ with µ(f,x) N the multiplicity of x as a zero of f. ∈ 0 ∪ {∞} The number of positive roots of f counted with multiplicities is at most the number of sign changes of the coefficient sequence of f and the number of negative roots of f counted with multiplicities is at most the number of sign changes of the coefficient sequence of f( X). If f is an RZ-polynomial, then in the last statements we even have equality. − Proof. The first statement is well-known (e.g. [Be]). If f is an RZ-polynomial we have bj +bj deg(f) j 1, ..., d 1 ab ab < 0 + j 1, ..., d 1 ( 1) +1 ab ab < ≥ |{ ∈ { − } | j j+1 }| |{ ∈ { − } | − j j+1 0 + b1 µ(f,x) + µ(f,x) + µ(f, 0) = deg(f); hence there must be }| ≥ x∈R>0 x∈R<0 equality everywhere. P P Corollary 5.3. Let (bj)j∈{1,...,d} be a strictly increasing tuple of natural numbers and 0, d bj f = ab X R[X] 0 be an RZ-polynomial such that ab R 0 for all j. Then j=1 j ∈ \{ } j ∈ \{ } for every j 1, ..., d 1 we have bj bj 2. If bj bj=2, then ab ab < 0. P ∈ { − } +1 − ≤ +1 − j j+1 Proof. The sign changes in the coefficient sequences f and f( X) complement each other well and it is easy to see that the claimed properties of the co−efficients of f are necessary in order to have a total of deg(f) b1 sign changes in the coefficient sequences of f and f( X). − − m 1 g Lemma 5.4. Let k 2. Assume fk = j=1 αj with αj R[ α,β, ..., α,β α, β ≥ m ∈ X X | ∈ 1, ..., k ] non-constant polynomials. Then also fk = αj g and those polyno- { } Q −1 j=1 |S (k−1) mials are non-constant. Here we identify Sg(k 1) as a subset of Sg(k) by looking at the − Q embedding ι : Sg(k 1) Sg(k), A A 0. There are other natural ways to embed − → 7→ ⊕ Sg(k 1) into Sg(k) (i.e. A U ∗(A 0)U for a unitary U Ck×k), however Lemma 5.1 states− that they lead to the same7→ results.⊕ ∈ 24 Proof. Write αj k := αj g . We only have to show that α k is not constant. So | −1 |S (k−1) 1| −1 assume α k = 1. Since α is invariant under unitary tranformations, we know that 1| −1 1 α1 is constant on all tuples of Hermitian matrices which share a kernel vector. We write r α1 = s=0 ps, where the ps are homogeneous of degree s. We have p0 = 1; thus the ps with s 1 have to vanish on the matrices which share a kernel vector. P≥ It is easy to see that a non-trivial linear polynomial cannot vanish on all those matrices g (since there span is all S (k)). Therefore p1 = 0. Now we claim that p2 switches its sign or is zero. Suppose p = 0. If p does not change the sign, it is up to sign a sum of squares 2 6 2 of linear forms. However with the same argument as before, this would imply p2 = 0. So p2 switches its sign. Thus we find an A Sg(k) with p (A) > 0 or p = 0. In any case the polynomial function ∈ 2 2 q : R R,t α (tA) cannot be an RZ-polynomial because of Corollary 5.3. → 7→ 1 Theorem 5.5. There is M N and N N such that each fk decomposes into a product ∈1 g ∈ of N polynomials gj,k R[ , ..., 1 α, β k] with gj,k(0) = 1. For k M they ∈ Xα,β Xα,β | ≤ ≤ ≥ are irreducible. We can achieve that gj,k g = gj,ℓ for 1 ℓ k and j 1, ..., N . |S (ℓ) ≤ ≤ ∈ { } Proof. The preceeding lemma tells us that there are N, M N guaranteeing that fk de- ∈ 1 g composes into a product of N irreducible polynomials g ,k, ..., gN,k R[ , ..., 1 1 ∈ Xα,β Xα,β | ≤ α, β k] for k M such that gj,k(0) = 1. For smaller k of course we have that fk ≤ ≥ decomposes into a product of Nk irreducible polynomials gj,k with Nk N. Set Nk = N ≥ for k M. Fix now k M and take ℓ < k. Write s = k ℓ. We define hj,ℓ = gj,k Sg(ℓ) ≥ ≥ g g − | and pj,s = gj,k g . Now for Y S (ℓ), Z S (s) we know that |S (s) ∈ ∈ Nℓ Ns gj,ℓ(Y ) gj,s(Z)= fℓ(Y )fs(Z)= fk(Y 0)fk(0 Z) ⊕ ⊕ j j Y=1 Y=1 N N = fk(Y Z)= fk(Y 0)fk(Z 0) = gj,k(Y 0)gj,k(Z 0) = hj,ℓ(Y )pj,s(Z) ⊕ ⊕ ⊕ ⊕ ⊕ j j Y=1 Y=1 Now suppose ℓ M. Then we know that Nℓ = N and from the equation above we get ≥ N N gj,ℓ(Y )= hj,ℓ(Y ) j j Y=1 Y=1 Since the prime factorization is unique and all the polynomials on the left are irreducible, the same factors appear on both sides. So we can in a unique way (up to multiplicities) rearrange the indices in such a way that gj,k Sg (ℓ) = gj,ℓ holds. Repeating this procedure for growing levels k and all M ℓ < k gives the| desired result. ≤ Corollary 5.6. Let L be a monic linear pencil and fix the gj,k, N, M from Theorem 5.5. Then for fixed j 1, ..., N the sequence of closures of the connected components of the g −1 ∈ { g } sets S (k) gj,k (0) S (k) around 0 form the levels of a closed matrix convex set Dj. We N \ ⊆ have Dj = L. j=1 D g −1 Proof.TFor j 1, ..., n let Ej(k) be the connected component of the set S (k) gj,k (0) g ∈ { } \ ⊆ S (k) around 0. We use Lemma 1.10 to verify that the Ej are matrix convex. As a factor of a real-zero polynomial every gj,k is a real-zero polynomial. Thus each Ej(k) is ordinary convex. As a determinant fk is unaffected by unitary conjugations. Therefore Lemma 5.1 implies that this is also the case for the gj,k. The closedness under taking direct sums and reducing subspaces follows from gj,k1 (Y )gj,k2 (Z)= gj,k1+k2 (Y Z) (see the calculation in 25 ⊕ the proof of Theorem 5.5). Hence Ej is matrix convex and also its closure Dj is. −1 L(k) is the closure of the connected component of (fk) (R 0 ) around 0. By convexity D g \{ } g this set equals A S (k) t [0, 1) : fk(A) > 0 and also Dj(k) = A S (k) t { ∈ | ∀ ∈N } { ∈ | ∀ ∈ [0, 1) : gj,k(A) > 0 . From this Dj = L follows directly. } j=1 D Lemma 5.7. Let f R[Y1, ..., YTn] be an RZ-polynomial and S the generated closed convex set. Suppose that there∈ is no other RZ-polynomial of lesser degree defining the same convex set. Then f is regular on a dense subset of ∂S. Proof. This follows from the fact that x Rn f(x) = 0 has locally dimension n 1 around every point x ∂S ([HV, Lemma{ 2.1])∈ and| [BCR, Definition} 3.3.4, Theorem 4.5.1− and Proposition 3.3.14].∈

Lemma 5.8. (cf. [HV, Lemma 2.1]) Let f, g R[Y1, ..., Yn], f be an irreducible RZ- ∈ −1 polynomial and Cf the connected component of f (R 0 ) around 0. Suppose that \ { } there is no other RZ-polynomial of lesser degree defining the same convex set Cf . Suppose there is a point x ∂Cf and ε> 0 such that g = 0 on ∂Cf B(x, ε). Then f divides g. ∈ ∩ Proof. From [HV, Lemma 2.1] and [BCR, Proposition 8.2.2] we know that ∂Cf B(x, ε) ∩ has dimension n 1 as a semialgebraic set and the R-Zariski closure of ∂Cf B(x, ε) has − ∩ dimension n 1 as a real variety. Since Z(f) := x Rn f(x) = 0 is irreducible either Z(f) Z(g)−or the intersection Z(f) Z(g) has dimension{ ∈ | smaller than} n 1. Therefore Z(f) ⊆ Z(g). Since (f) is real [BCR,∩ Theorem 4.5.1], f divides g. − ⊆ 5.2. Gleichstellensatz. a In this chapter and the following we want to analyze the following questions: Given a free spectrahedron S = L, can we find an easier and in some sense minimal description of S as a spectrahedron?D How does this minimal description relate to the monic linear pencil L and is it unique? We will show that there are in some sense "minimal and indecomposable" pencils which form the building blocks to construct every spectrahedron containing 0 in its interior and also every monic linear pencil. The decomposition into those building blocks will be unique up to unitary equivalence. In the literature there are different approaches to deal with these questions. The most obvious way to approach these questions is to look at the spectrahedron L directly. g D We call L S irreducible if the L1, ..., Lg admit no joint non-trivial invariant subspace. Helton, Klep∈ and McCullough in [HKM4] call the tuple L minimal if no smaller tuple H describes the spectrahedron L. They showed that two minimal descriptions of the same bounded spectrahedron areD unitary equivalent (Gleichstellensatz). Furthermore they showed that the minimal tuple H defining L is a direct summand of L. We will give a new proof of these results and generalizeD them to the unbounded setting (which was established already by Zalar in [Z] by improving the techniques of [HKM4]). Furthermore we will show how these results lead to the existence of the "irreducible" building blocks mentioned above. An alternative approach of Klep and Volˇciˇc in [KV] deals with the free locus (L) := X g Z { ∈ S ker L(X) = 0 instead of the spectrahedron L. This leads to the same building blocks.| We look6 further{ }} into this approach in the nextD subsection.

Definition 5.9. Let L be a monic linear pencil. We call L (or L) irreducible if L1, ..., Lg admit no joint non-trivial invariant subspace. We call L (or L) -irreducible if for all other monic linear pencils H , H the equality D 1 2 H1 H2 = L implies L = Hi for some i 1, 2 . D ∩ D D D D 26 ∈ { } We call L (or L) -minimal if there is no monic linear pencil H of smaller size such that D H = L. D D For the following we need a special case of the boundary theorem of Arveson. We will outline the elementary proof found by Davidson, adapted to our special situation. g n×n Lemma 5.10. ([D2]) Let L S (n) irreducible. Suppose that there are Cj C with r ∗ ∈ r ∗ r ∗ ∈ L = C LCj and I C Cj. Then C Cj = I and all Cj are scalar j=1 j j=1 j j=1 j multiples of the identity. P P P r ∗ r ∗ Proof. ([D2]) Consider C = I C Cj, J = 0 Ir. Then C Cj = In. Set 0 − j=1 j ⊕ j=0 j ∗ ∗ ∗ ∗ ∗ ∗ V = (C0 C1 ... Cr ). ThenqV is an isometry, i.e. V V = I, VV is the projection ∗ P n×n n×n P r ∗ onto im(V ), and V is contractive. We define ϕ : C C , A j=1 Cj ACj = ∗ → 7→ V (J A)V . Denote by S the operator system generated by the Li. ⊗ P n Now let W be a minimal non-zero subspace of C such that W is Cj-invariant for all n×n j 0, ..., m . Define Γ= D C j 0, ..., r w W : CjDw = DCjw = D ∈ { } { ∈ | ∀ ∈ { }∀ ∈ } { ∈ Cn×n w W : ((1 I) D)V w = V Dw . We show that if D S and B Γ, then DB |Γ ∀. ∈ ⊕ ⊗ } ∈ ∈ ∈ So take such D,B and set M = w W DB W w = DBw . Let w M. Then { ∈ | || | || || || || ||} ∈ DBw 2 = ϕ(D)Bw 2 = V ∗(J D)V Bw 2 = V ∗(J D)((1 I) B)V 2 || || || || || ⊗ || || ⊗ ⊕ ⊗ || r ∗ 2 2 2 = V (J DB)V w (J DB)V w = DBCjw || ⊗ || ≤ || ⊗ || || || j X=1 r 2 2 2 2 2 DB W Cjw DB W w = DBw ≤ || | || || || ≤ || | || || || || || j X=1 Therefore Cjw M for j 1, ..., m and C0w = 0. So we verified that CjM M for all j. By minimality∈ of W we∈ get { M = }W . We also have ⊆ (J DB)V w = V ∗(J DB)V w = VV ∗(J DB)V w || ⊗ || || ⊗ || || ⊗ || where the first equality follows from the calculation above and the second from the fact that V is an isometry. Hence ((1 I) DB)V w = (J DB)V w = VV ∗(J D)((1 I) B)V w = VV ∗(J D)V Bw ⊕ ⊗ ⊗ ⊗ ⊕ ⊗ ⊗ = V ϕ(D)Bw = V DBw where the second equality comes from the Pythagoras theorem and the first from C0w = 0. Thus DB Γ. Of course I Γ and we conclude that Γ contains the algebra Cn×n generated ∈ n∈×n n n×n by the Li. Now let D C , x C . Choose w W and B C such that Bw = x. We know ∈ ∈ ∈ ∈ Dx = DBw = V ∗V DBw = V ∗((1 I) DB)V w = V ∗(J DB)V w ⊗ ⊗ ⊗ = V ∗(J D)((I 1) B)V w = V ∗(J D)V Bw = ϕ(D)x ⊗ ⊕ ⊗ ⊗ where we used that DB,B Γ. We conclude ϕ(D) = D. We have ϕ(I) = I. Thus n ∈ ∗ ∗ ∗ ∗ C0 = 0. Now let v C with v = 1. With Cauchy-Schwarz 1= v vv v = v ϕ(vv )v = r ∗ ∗ ∗ ∈ r || || 2 r ∗ (v Cjv) (v Cjv) = v,Cj v Cjv,Cjv = v ϕ(I)v = 1 and there- j=1 j=1 |h i| ≤ j=1 |h i| fore span v is an invariant subspace of Cj. Thus Cj is a multiple of the identity. P P P Lemma 5.11. ([HKM4, Corollary 3.18] and [Z, Theorem 3.1]) Let L Sg(k) and H g ∈ ∈ S (h) be irreducible such that L = H. Then L and H are unitarily equivalent. D D 27 r ∗ Proof. We know that their exist Cj and Dℓ such that L = j=1 Cj HCj and H = s ∗ r ∗ s ∗ ℓ=1 Dℓ LDℓ with Ik j=1 Cj Cj and Ih ℓ=1 Dℓ Dℓ. Of course we can assume that ∗ ∗ rP s ∗ ∗ Cj Cj = 0 for all j and Dℓ Dℓ = 0 for all ℓ. We see that L = j=1 ℓ=1 Cj Dℓ LDℓCj as P 6 r s P∗ ∗6 P well as Ik j=1 ℓ=1 Cj Dℓ IkDℓCj. We apply Lemma 5.10 and find λℓ,j C such that r s ∗ ∗ P P ∈ r ∗ λℓ,jI = DℓCj. Additionally Ik = j=1 ℓ=1 Cj Dℓ IkDℓCj which implies Ik = j=1 Cj Cj. P ∗P ∗ s ∗ Fix j. Since 0 = C Cj = C ( D Dℓ)Cj, we know that λℓ,j = 0 for one ℓ. In particular 6 j j ℓ=1P ℓ P 6 P Cj is injective and k h. ≤ P The same calculation the other way around shows that all Dℓ are also injective, Ih = s ∗ D Dℓ as well as h k and thus h = k. Since all Dℓ,Cj are injective, we have ℓ=1 ℓ ≤ that DℓCj = CjDℓ are a nonzero multiple of the identity. In particular all Cj are linear P dependent, so we can find λj = 0 and C such that Cj = λjC for all j. We conclude 6 ∗ r r r 2 ∗ 2 2 L = λj C HC = λj C H λj C. | | v | |  v | | j=1 uj=1 uj=1 X uX uX ∗ t  t r 2 r 2 Since I = λj C λj C we conclude that L and H are unitarily equiv- j=1 | | j=1 | | alent. qP qP Lemma 5.12. Let L be a monic linear pencil. Suppose det Lk = g ,k ... gN,k is the 1 · · decomposition of Corollary 5.6, where gj,k becomes eventually irreducible for big k and Dj is the induced closed matrix convex set. Let J 1, ..., N such that L = j∈J Dj and no matrix convex set on the right sight can be omitted⊆ { without} destroyingD equality. Then T for all j J the matrix convex set Dj is in fact a spectrahedron. ∈ ◦ Proof. For a tuple (Ki)i∈I of closed matrix convex sets containing 0 we have ( i∈I Ki) = K◦ and by the bipolarity theorem mconv( K ) = (( K )◦)◦ = ( K◦)◦. i∈I i i∈I i i∈I i Si∈I i TWe know from Corollary 2.12 that S S T ◦ ◦ ◦ ◦ mconv(L, 0) = ( L) = Dj = mconv D = mconv D . D    j   j  j\∈J j[∈J j[∈J    ◦    where the last equality comes from the fact that j∈J Dj is compact together with the free g Caratheodory theorem Lemma 1.8 and Proposition 3.4. Therefore we find (Lj)j∈J S ◦ S ◦ ⊆ with Lj D such that L mconv( Lj j J ). From Lj D it is clear that ∈ j ∈ { | ∈ } ∈ j L Dj L. We also have L = L . We claim L = Dj for all j J. D j ⊇ ⊇ D D j∈J D j D j ∈ Fix j J and choose M such that gj,kT is irreducible and Dh(k) = L(k) for all ∈ h∈J\{j} 6 D k > M. Then we know gj,k detk Lj from Lemma 5.8 which implies the other inclusion | T L Dj. D j ⊆ Proposition 5.13. Let L Sg(k) be -irreducible and -minimal. Then L is irreducible. ∈ D D

Theorem 5.14. Let L , ..., Ls be -irreducible and -minimal such that the Zariski clo- 1 D D sure of each ∂DLj (k) is irreducible in the Zariski topology for big k. Suppose that s s L = L . D j 6 D j j \=1 j=1\,j6=ℓ g s for all ℓ 1, ..., s . Suppose L S (δ) deﬁnes a spectrahedron L = L . Then up ∈ { } ∈ D j=1 D j to unitary equivalence L1 ... Ls is a direct summand of L. ⊕ ⊕ 28 T Proof. From Theorem X.19 and Corollary X.15 we know that we can write s r s r ∗ ∗ ∗ L = Vj,ℓLjVj,ℓ and I = WW + Vj,ℓVj,ℓ. j j X=1 Xℓ=1 X=1 Xℓ=1

For k N denote Tj(k)= ∂ L (k) int L (k) . Choose k N in such a way that ∈ D j ∩ h6=j D h 0 ∈ the Zariski closure of each ∂ (k) is irreducible and T (k) = for k k . Let k k . Lj T j 0 0 D δ k 6 ∅ ≥ δ ≥ Suppose Xh Th(k) and vh C C such that L(Xh)vh = 0. Write vh = α=1 eα vh,α ∈ k ∈ ⊗ ∗ ⊗ with vh,α C . Now since Lj(Xh) 0 for j = h, we conclude that 0 = (Vj,ℓ Iδ)vh, ∗ ∈ ≻ 6 P ⊗ 0 = (W Iδ)vh. Denote δh = dim M(Xh, vh)L. ⊗ h Now choose an orthonormal basis wh,1, ..., wh,δh of M(Xh, vh)L. Let Uh(Xh, vh) = U δ×δ ∈ C h be the matrix where the β-th row contains the coeﬃcients of vh,β modulo the basis h wh,1, ..., wh,δh . Then U has rank δh. For ﬁxed α we have:

δh δh h h eα vh,α = eα U wh,γ = U eα wh,γ ⊗ ⊗ α,γ α,γ ⊗ γ=1 γ=1 X X We calculate for h = j 6 δh r δ r δ r ∗ h ∗ ∗ V U eα wh,γ = (V I) (eα vh,α)= (V I)vh = 0 j,ℓ α,γ ⊗ j,ℓ ⊗ ⊗ j,ℓ ⊗ γ=1 α=1 ! α=1 X Xℓ=1 X Xℓ=1 X Xℓ=1 Therefore for all γ 1, ..., δh the equality ∈ { } r δ r ∗ h ∗ h 0= Vj,ℓ Uα,γ eα = Vj,ℓU·,γ α Xℓ=1 X=1 Xℓ=1 holds, so r V ∗ U h = 0 for j = h. Analogeously W ∗U h = 0. Because of WW ∗ + ℓ=1 j,ℓ 6 s r V V ∗ = I we have r V V ∗ U h = U h. This means that r V V ∗ j=1 ℓ=1P j,ℓ j,ℓ δ ℓ=1 h,ℓ h,ℓ ℓ=1 h,ℓ h,ℓ is the identity on the δ -dimensional space im(U (X , v )). Since the eigenspaces of a P P h P h h h P Hermitian matrix are pairwise orthogonal, we get the following: δ k Set Sj = span im(Uj(X, v)) k k0, X Tj(k), v C C : L(X)v = 0 for all j, { | ≥ ∃ ∈ ∗ ∈ ⊗ } where Uj is deﬁned in the same way as Uh. Then Vj,ℓ Sh = 0 and Sh Sj for j = h. We δ s | ⊥ ∗ 6 ∗ decompose C = j=1 Sj T . With respect to this decomposition: Write Yj,ℓ := Vj,ℓ Sj , ∗ ∗ ⊕ | Z := V T j,ℓ j,ℓ| L ∗ ∗ Y1,ℓL1Y1,ℓ ... 0 Y1,ℓL1Z1,ℓ . .. . . L = P . . . P .  ∗ ∗ 0 ... Ys,ℓLsYs,ℓ Ys,ℓLsZs,ℓ  ∗ ∗ ∗   Z ,ℓL Y ... Zs,ℓLsY Zj,ℓLjZ   1 1 1,ℓ P s,ℓ P j,ℓ  ∗ ∗  P Y1,ℓY1,ℓ P... 0 PP Y1,ℓZ1,ℓ . .. . . I = P . . . P .  ∗ ∗ 0 ... Ys,ℓY Ys,ℓZ  s,ℓ s,ℓ   Z Y ∗ ... Z Y ∗ Z Z∗   1,ℓ 1,ℓ P s,ℓ s,ℓ P j,ℓ j,ℓ   Fix j 1, ..., s . Let P be theP projection ontoPSj. We wantPP to determine P LP ∗ . Let ∈ { } δ k D Xj Tj(k) and vj C C with L(Xj)vj = 0. With the same notation as above ∈ ∈ ⊗ 29 j (U = Uj(Xj, vj)) we see

δj δj δj j j j vj = eα vj,α = eα U wj,γ = U eα wj,γ = U wj,γ ⊗ ⊗ α,γ α,γ ⊗ ·,γ ⊗ α α γ=1 α γ=1 γ=1 X X X X X X δj j 0 = (P I)0 = (P I)L(Xj)vj = (P I)L(Xj ) U wj,γ ⊗ ⊗ ⊗  ·,γ ⊗  γ X=1 δj   ∗ j ∗ = (P I)L(Xj)(P I) U wj,γ = (P LP )(Xj )vj ⊗ ⊗  ·,γ ⊗  γ=1 X ∗ We conclude P LP ∗ L with Lemma 5.8. The destription of P LP in terms of Lj D ⊆ D j ∗ ∗ yields that P LP Lj ; hence we have P LP = Lj . Lemma 5.11 in connection with D ⊇ D D D ∗ Proposition 5.13 and the fact that Lj is -irreducible imply that Lj is a submatrix of P LP D and L. That Lj is even a direct summand is implied by the following easy proposition. L B Proposition 5.15. Let = r V ∗LV with r V ∗V = I and L be irre- B∗ C j=1 j j j=1 j j ducible. Then B = 0. P P ∗ ∗ r Wj Wj Wj Zj I 0 Proof. Write Vj = Wj Zj . Then we know j=1 ∗ ∗ = . Because Zj Wj Zj Zj 0 I r ∗ of Lemma 5.10 and L = W LWj we knowP that there exist λj C such that Wj = j=1 j ∈ λ I. In particular we have B = r W ∗LZ = L r λ∗Z = L r W ∗Z = 0. j P j=1 j j j=1 j j j=1 j j Corollary 5.16. Let L be a monicP pencil. Then LPis a direct sumP of monic linear pencils Lj which are simultaneously -minimal and -irreducible (and constant pencils 1) and D D such that the Zariski closure of ∂ L (k) is irreducible for k big enough. D j Proof. We prove the claim by induction on size(L). The case size(L) = 1 is clear. Suppose that size(L) > 1, L has no constant direct summand and det Lk = g ,k ... gM,k is the 1 · · decomposition of Corollary 5.6, where gj,k becomes eventually irreducible for big k and Dj is the induced closed matrix convex set. Let J 1, ..., M such that L = j∈J Dj and no matrix convex set on the right sight can be omitted⊆ { without} destroyingD equality. Then T Lemma 5.12 allows us to write Dj = L with Lj -minimal. Since the Zariski closure of D j D ∂ L (k) is irreducible for big k, we know that Lj is also -irreducible. Theorem 5.14 tells D j D us that Lj is a direct summand of L. Thus we can write L Lj H. Now j∈J ≈ j∈J ⊕ apply the induction hypothesis to H. L L Finally, we can formulate our version of the Gleichstellensatz. Originally Helton, Klep and McCullough proved the Gleichstellensatz using the Silov ideal of an operator system, whose existence is already a deep result in operator theory. Since they need the Silov boundary only for operator systems living in the set of bounded linear operators of a fixed finite-dimensional Hilbert space, it could be possible that the proof of the existence of the Silov ideal in that case can be carried out in a more elementary way. Corollary 5.17. (Gleichstellensatz) (cf. [HKM4, Corollary 3.18] and [Z, Theorem 3.1]) Let L be a monic linear pencil and S = L. Then there exist monic linear pencils L , ..., Lr [Example 7.3, D 1 satisfying the following: Each Lj is -irreducible and -minimal. For each j 1, ..., r Example 7.4, D D ∈ { } and big k the Zariski closure of ∂ Lj (k) h6=j int Lh (k) is non-empty and irreducible. Example 7.5] r D ∩ D Lj is -minimal and a direct summand of all monic linear pencils H with S = H. j=1 D T D The Lj are uniquely determined up to order and unitary equivalence. L Proof. Combine Theorem 5.14 and Lemma 5.12. 30 5.3. Characterisation of -irreducible and -minimal pencils. D D Definition 5.18. For a monic linear pencil L we define the free locus (L) := X Z { ∈ Sg ker L(X) = 0 . We call a monic linear pencil L (or L) -irreducible if for all other | 6 { }} Z monic linear pencils L1, L2 the equality (L1) (L2)= Z(L) implies (L)= (Lj) for some j 1, 2 . Z ∪Z Z Z ∈ { } We call a monic linear pencil L (or L) -minimal if there is no monic linear pencil H of smaller size such that (L)= (H). Z Z Z g ∗ Let L S (k). Then L (or L) shall denote the unital C -algebra generated by L , ..., Lg. ∈ A A 1 -irreducible pencils were introduced by Klep and Volˇciˇc in [KV] in a more general setting (allowingZ also non-Hermitian pencils). We recall their main result: Theorem 5.19. [KV, Theorem 5.4] Let L,H Sg. Then (L) (H) if and only ∗ ∈ Z ⊆ Z if there is a -homomorphism of C -algebras ϕ : H L such that Li = ϕ(Hi) for all i 1, ...,∗ g . In case that (L) = (H), theA map→ϕ Ais even an isomorphism. L is ∈ { } Z Z -irreducible if and only if L is simple. Z A Theorem 5.20. Let L be a non-constant monic linear pencil of size δ. Then the following is equivalent: (a) L is -irreducible and -minimal (b) L is Z-irreducible and Z-minimal D D (c) pz( L)= δ D δ×δ (d) L is irreducible (i.e. L = C ). A (e) There is k N such that the for all k N k the determinant detk L is irreducible. 0 ∈ ∈ ≥ Proof. (a) (e): This is a direct consequence of Corollary 5.26, which we will prove in the rest of⇐⇒ the chapter. (b) = (a): Suppose L is -irreducible and -minimal. Obviously L is -minimal. Now ⇒ D D Z take monic linear pencils L ,L such that (L ) (L )= (L). If L , L = L, then 1 2 Z 1 ∪Z 2 Z D 1 D 2 6 D L was not -irreducible. So suppose L1 = L. Then the Gleichstellensatz implies that L is a submatrixD of L . Thus (L )=D (L).D 1 Z 1 Z (a) = (b): Suppose L is -irreducible and -minimal. We know that L decomposes ⇒ Z Z into a direct sum of -irreducible and -minimal pencils Lj. Due to (b) = (a) they are also -irreducibleD and -minimal. SoD the decomposition has been trivial.⇒ Z Z (b) (d): This follows directly from Corollary 5.17. ⇐⇒ (c) = (b): Suppose pz( L)= δ. If L is not -minimal, then there exists a monic linear ⇒ D D pencil H with size less than pz( H). This contradicts Corollary 3.3. So L is -minimal. Assume L is not -irreducible. ByD Corollary 5.16 we know that L is a direct sumD of smaller -irreducible pencils.D The pz-number of L is less than the maximum of the pz-number of Dthe summands. So there cannot be more than one summand.

(b) = (c): Suppose now that L is -irreducible and -minimal. Assume that pz( L) < ⇒ D D D δ. Corollary 3.9 and Corollary 5.16 allow us to ﬁnd a monic linear pencil H = H ... Hs 1 ⊕ ⊕ such that H = L, where each Hj has size smaller than δ. Since L is -irreducible, we can supposeD s =D 1. Since L is also -minimal,δ > size(H ) is a contradiction.D D 1 Proposition 5.21. A monic linear pencil L is -irreducible if and only if for all pencils Z L , L with (L) (L ) (L ) we have (L) (Lj) for some j. 1 2 Z ⊆Z 1 ∪Z 2 Z ⊆Z Proof. Let L be -irreducible. Theorem 5.19 says that L is simple. Suppose (L) Z A ∗ Z ⊆ (L1) (L2) holds. From Theorem 5.19 we know that there is a unital -homomorphism Z ∪Z 31 ϕ : L L L induced by ((L )i, (L )i) Li. Since L is simple, ϕ is surjective. A 1 ×A 2 → A 1 2 7→ A The ideals of L L are direct products of ideals of the L . Now we divide by the kernel A 1 ×A 2 A i of ϕ to see that there are ideals Ij of Lj for j 1, 2 such that ϕ : ( L1 L2 )/(I1 I2) ∗ A ∈ { } A ×A × → L is a C -isomorphism. The domain is isomorphic to ( L1 /I1) ( L2 /I2) and simple. A A ∗ × A Thus there is an j such that L = Ij and ϕ induces also a C -homomorphism from one A j of the L to L. Another usage of Theorem 5.19 yields the result. A k A The upcoming Corollary 5.24 characterizes -irreducible linear pencils L as those whose determinants are irreducible polynomial at highZ levels. In a similar formulation Klep and Volˇciˇc announced a proof in [KV] for a later paper. Using completely diﬀerent techniques than ours, they gave a proof in [HKV]. We do not think that our proof generalizes to the non-Hermitian setting in [HKV].

β1 βN Lemma 5.22. Let L be a monic linear pencil of size δ. Suppose detk L = g ... g 1,k · · N,k is the decomposition as in Corollary 5.6, where βj N denote multiplicities and such that ∈ for k M the polynomials gj,k are irreducible and pairwise not associated and Dj is the ≥ induced closed matrix convex set. Then each pz(Dj) δ and Dj is a spectrahedron for all j 1, ..., N . ≤ ∈ { } on Proof. Let fk := detk L. We denote by ∂Dj(k) = A ∂Dj(k) h 1, ..., N j : { ∈ | ∀ ∈ { } \ { } gh,k(B) = 0 . Lemma 5.8 tells us that those points are dense in ∂Dj (k) for k > M in the 6 } on Euclidean topology. So fix j and let X ∂Dj(k) with k M, so gj,k(X) = 0. ∈ ≥ δ δ k k Choose v = eα vα C C with L(X)v = 0. Let P be the projection of C onto α=1 ⊗ ∈ ⊗ M(X, v)L = span(v1, ..., vδ ). We calculate: P g δ 0= L(X)v = I I + Li Xi (eα vα)  ⊗ ⊗  ⊗ j α X=1 X=1 g δ  ∗ ∗ = I I + Li Xi (I P )(eα vα)= L(XP )v  ⊗ ⊗  ⊗ ⊗ j α=1 X=1 X  ∗  ∗ ∗ ∗ and therefore L(P XP )v = (I P )L(XP )v = 0. Set Xα = (1 α)P [P XP ]P + αX ⊗ − for α [0, 1]. The previous calculations mean that L(Xα)v = 0. Thus fk(Xα) = 0. ∈ We conclude that gj,k(Xα) = 0 for α near 1 because gh,k(X) = 0 for h = j. Since ∗ ∗ 6 6 ∗ gj,k is a polynomial gj,k(X0) = 0. Thus 0 = gj,k(P [P XP ]P ) = gj,k(P XP 0) = ∗ ∗ ⊕ gj,rk(P )(P XP ). On the other hand we have X Dj; thus also P XP Dj by matrix convexity. ∈ ∈ on We have checked that for every X ∂Dj(k) with k M there exists a projection P ∗∈ ≥ of rank at most δ such that P XP ∂Dj. Now let A ∂Dj(k) with k N. Choose ∈ ′ ∈ ∈ s N such that s + k M. Then A = A (Is 0) ∂Dj(s + k). Approximate ′∈ ≥ on ⊕ ⊗ ∈ A by points of ∂Dj (k + s) . With the help of Bolzano-Weierstraß we get a projection k+s ′ ∗ ′ ∗ P : C im(P ) of rank at most δ such that P A P ∂Dj. We have P A P → ∈ ∈ mconv(A, 0)(δ) = mconv( QAQ∗ Q : Ck im(Q) projection of rank at most δ 0 ) { | → } ∪ { } (Proposition 3.4). Moreover 0 int(Dj), so there is one projection Q of rank at most δ ∗ ∈ such that Q AQ ∂Dj. This means pz(Dj) δ. ∈ ≤ Now we imitate our proof that "closures of matrix convex free basic open semialgebraic" sets are spectrahedra to show that each Dj is a spectrahedron. The reason why this is on possible is that we did not merely prove pz(Dj) δ, but also that for a point X ∂Dj(k) ∗ ≤ ∈ with L(X)v = 0 we have P AP ∂Dj where P is the projection onto M(X, v)L. Let ∈ 32 ε = max δ, M . { } on on We work with ∂Dj(ε) . We consider one Nash manifold C of ∂Dj(ε) with associated Nash function fC that maps each A C to a non-trivial kernel vector of L(A). Fix a dense ∈ ε subset (Xn)n of C and write fC (Xn) = (fC, (Xn), ..., fC,δ(Xn)) with fC,j(Xn) C . ∈N 1 ∈ We apply the techniques of the proof of Lemma 4.7 and Theorem 4.2 and achieve: There δ×δ δ×δ exist monic H := HC SC X , V SC such that HC (Xn)V (Xn)fC (Xn) = 0 for all ∈ h i1 ∈ n N, V (Xm)fC (Xm) = 0 for one m N as well as H(X) 0 for all X int(Dj). ∈ 6 ∈ ≻ ∈ δε Therefore we conclude that the function φ : C C , X V (X)fC (X) is not zero and → 7→ H(X)V (X)fC (X) = 0 for all X C. Since C is a Nash manifold and V is constant, we know that V is in each component∈ a Nash function on C. Without loss of generality we can assume that C is a relatively open box. But then Remark 4.9 implies that φ−1(Cδε 0 ) is dense in C and therefore H(X) has \ { } non-trivial kernel for all X C. This means that H separates all points of C from int(Dj). ∈ on Therefore the direct sum of all HC separates all points of ∂Sj(ε) from int(Dj). Due to on the Zariski density of ∂Dj(ε) in ∂Dj (ε) we get that the direct sum of all HC separates all points of ∂Sj(ε) from int(Dj). Since pz(Dj) δ, pz( H ) δ and the spectrahedron ≤ D C ≤ defined by the direct sum of the HC coincides with Dj on level ε, the claim follows. Theorem 5.23. (cf. [HKV, Theorem 3.4]) Let L be a monic linear pencil. Then L is n -irreducible if and only for big k we have detk L = qk where qk is irreducible and n does notZ depend on k. Proof. " =": This is obvious. ⇐ α1 αn "= ": Suppose det Lk = g ... g is the decomposition as in Corollary 5.6, where gj,k ⇒ 1,k · · N,k eventually becomes irreducible for big j, Dj is the induced closed matrix convex set and the gj,k are pairwise not associated for all big fixed k. Every Dj is a spectrahedron, so write Dj = L with Lj -minimal. It is clear that the Lj are also -irreducible. Hence D j D D the Lj are also -irreducible and -minimal. We distinguish two cases: Z Z Case 1: There is no j such that (Lj) = (L). However (L) (Lj). This Z Z Z ⊆ j Z contradicts the fact that L is -irreducible (Proposition 5.21). Z S Case 2: There is an j such (Lj) = (L). WLOG assume that the vanishing ideal of Z Z the Zariski closure of ∂ L is generated by g ,k (the Zariski closure must be irreducible D 1 otherwise we are in case 1) and that (L )= (L). Since g ,k is not associated to g ,k we Z 1 Z 2 1 have L L . D 2 ⊃ D 1 We want that (L2) (L1). If this is not the case, again we apply Lemma 5.22 to L2 Z β⊂1 Z βM′ ′ and write detk L = h ... h ′ with associated spectrahedra E (s 1, ..., M ) where 2 1,k · · M ,k D s ∈ { } h1,k = g2,k and all other hℓ,k are not asssociated to h1,k. Then we have (L2) (L1) M Z ⊆Z ∪ (Es). L is -irreducible and (L ) * (Eℓ) for ℓ 2, thus (L ) (L ). s=2 Z 2 Z Z 2 Z ≥ Z 2 ⊂Z 1 ∗ Theorem 5.19 says that the mapping rule (L )i (L )i defines a unital -homomorphism S 1 7→ 2 ϕ : L1 L2 . However L1 and L2 are simple so there is some unitary U such that L =AU −→1L AU (Lemma X.5A). HenceA (L )= (L ), a contradiction. 1 2 Z 2 Z 1 Corollary 5.24. (cf. [HKV, Theorem 3.4]) Let L be a monic linear pencil such that L is -minimal and -irreducible. Then for k big enough detk L is irreducible. D D Proof. We have δ := size(L) = pz( L) due to Theorem 5.20. From Theorem 5.23 we know D already that detk L is a power of an irreducible polynomial for large k. Hence we are 33 finished if we can show that there is some X ∂ L(δ) such that the kernel of L(X) is ∈ D one-dimensional. Indeed, then detδ L cannot have a double zero in X. δ Since pz( L) = δ, we find X ∂ L(δ) such that for each projection P of C onto a D ∗ ∈ D δ δ smaller space L(P XP ) 0. Now assume v = eα vα and w = eα wα are ≻ α=1 ⊗ α=1 ⊗ linearly indepent members of ker L(X). By construction we know that dim M(X, v)L = P P dim M(X, w)L = δ. Consider the invertible linearly independent matrices V = (v1 ... vδ), W = (w ... wδ). Then the polynomial det(V + tW ) is not constant and has a root λ C. 1 ∈ Then v + λw ker L(X) 0 , however M(X, v + λw)L has dimension less than δ. ∈ \ { } Remark 5.25. The preceding proof shows: If = C BX is a linear pencil of size δ, ∗ L − δ X L(δ) such that (P XP ) 0 for all projections P : C im(P ) of rank at most ∈ D L 2 ≻ → δ 1, then there exists v δ −1 such that dim M(X, v) = δ and ker( (X)) = span v . − ∈ S L L { } β1 βN Corollary 5.26. Let L be a monic linear pencil of size δ. Suppose detk L = g ... g 1,k · · N,k is the decomposition as in Corollary 5.6, where βj N denote multiplicities and such that ∈ for k M the polynomials gj,k are irreducible and pairwise not associated and Dj is the induced≥ closed matrix convex set. Then for all j 1, ..., N there is a simultaneously ∈ { } N -irreducible and -minimal monic linear pencil Lj such that L (Iβ ) Lj. D D ≈ j=1 j ⊗ r Proof. Write L as a direct sum L = Lj of -irreducible and L-minimal pencils Lj j=1 D D and observe that det L = r det L . k j=1 k Lj Remark 5.27. Lemma 5.4Qshowed that the number of irreducible factors of the determinant of a monic linear pencil regarded as a commutative polynomial on a fixed level k is decreasing as a function of k. We are going to justify why this function is not always constant. In [Br] P. Brändén has given an example of an RZ-polynomial p such that no power of f can be written as the level 1-determinant of a monic linear pencil but such that an RZ-polynomial q exists such that the connected component of q−1(R 0 ) around 0 is con- \{ } taining the connected component of p−1(R 0 ) around 0 and pq is the level 1-determinant \{ } of a monic linear pencil L. Set fk = detk L( ) X Write p = p1...ps and q = q1...qt as products of irreducible polynomials. Assume now that each fk was a product of s + t irreducible polynomials. Then we could write fk = 1 s 1 t j j i i i i j j pk...pkqk...qk with pk and qk irreducible such that p1 = p , q1 = q and pk Sg (ℓ) = pℓ, j j | qk Sg(ℓ) = qℓ for ℓ k. Now Corollary 5.26 in connection with Corollary 5.24 would imply | i i ≤ all the p1 and q1 are level 1-determinants of monic linear pencils. The generalized Lax-conjecture is equivalent to the fact that for every RZ-polynomial p with associated convex set Cp we can find a RZ-polynomial q as above such that Cp Cq and pq can be written as the level 1 determinant of a monic linear pencil (see [Br]). ⊆ In case that p is an irreducible RZ-polynomial that cannot be written as the level 1 determinant of a monic linear pencil and q is a RZ-polynomial of minimal degree such that pq = det1 L( ) and Cp Cq, the results from this chapter imply that detk L( ) is irreducible for bigX k. ⊆ X

6. Notions of extreme points for matrix convex sets In the classical case of a compact convex set T in Rn, the extreme points of T form the "minimal" set of points whose convex hull is T (Minkowski theorem). In the infinite- dimensional setting the Krein-Milman gives a similar statement involving closures. Let 34 K Sg be a compact matrix convex set. We would like to find a similar notion of extreme points⊆ which generalizes the Minkowski/Krein-Milman theorem to the matrix convex setting. Taking ordinary extreme points at every level is clearly too much (for instance when kz(K) < ). Also they dont reflect the nature of matrix convex combinations. Farenick, Morenz, Webster∞ and Winkler have developed a theory of matrix extreme points. In this setting the first half of the Krein-Milman theorem holds in the sense that every compact matrix convex set K has enough matrix extreme points to reconstruct itself, how the set of matrix extreme points does not need to the "minimal" set of generators ([WW], since minimality is an issue we call their theorem "weak" (free) Krein-Milman theorem). The setting in [WW] is more general since we consider only matrix convex sets consisting of g-tuples of matrices where g < while Webster and Winkler allow tuples of matrices of infinite size. However for most applications∞ the less general setting is enough. We will give a new and easier proof of the weak free Krein-Milman in our finite-dimensional setting (in the appendix we explain another variant of that proof). Indeed we will even show the free analogues of the first half of the classical Minkowski theorem and the Straszewicz theorem, which are both stronger than the Krein-Milman theorem. On the downside, the set of the matrix extreme points is only in some very weak sense a minimal generator of K. Also the notion of matrix extreme points does not really use the full free setting since one restricts the view only to finitely many levels. Another notion called absolute extreme point was introduced in [EHKM] by Evert, Helton, Klep and McCullough. This smaller set of points makes use of all the levels; however it is not clear in which cases the related first half of the Krein-Milman-Theorem holds. We will even see examples of compact matrix convex sets which do not have any absolute extreme points. For K being the polar of a spectrahedron, it was proven in [EHKM] that the set of absolute extreme points is finite (up to unitary equivalence) and the smallest set of points generating K as a matrix convex set. We will generalize this theorem to the case where K has finite kz-number and is compact. The main result (general Gleichstellensatz/strong free Krein-Milman) of this chapter will be a perfect analogue of the classical Krein-Milman g theorem. It states that there is a smallest operator tuple L h( ) defining K in the sense of K = mconv(L). This tuple L features all the absolute∈ B H extreme points and a generalized version of absolute extreme points. For this chapter we borrow the notation of matrix extreme and absolute extreme points from [EHKM]. 6.1. Generating matrix convex compact sets by different kinds of extreme points. Definition and Proposition 6.1. Let K Sg be a matrix convex set and A K(δ). We call A an (ordinary) extreme point of K⊆ if A is an extreme point of the convex∈ set K(δ) (where we treat Sg(δ) as a real vector space). r ∗ We call A matrix extreme if for all matrix convex combinations A = j=1 Vj BjVj kj ×δ r ∗ where Bj K(kj) and the Vj C are surjective and V Vj = Iδ, we already ∈ ∈ j=1 j P have that each kj = δ and A Bj. One can weaken the hypothesis of Vj surjective to be ≈ P kj δ and Vj = 0; also (at the same time) one can strengthen the conclusion that there ≤ δ×δ 6 r ∗ ∗ are Uj C unitary and λ C with λ = 1 such that Bj = U AUj and Vj = λjU . ∈ ∈ || ||2 j j mext(K) shall denote the set of matrix extreme points of K. r ∗ We call A absolute extreme if for all matrix convex combinations A = j=1 Vj BjVj where B K(k ), V Ckj ×δ 0 and r V ∗V = I we already have k δ and j j j j=1 j j δ P j ∈ ∈ \ { } 35 ≥ P g k ×k there is Cj S (kj δ) such that A Cj Bj. In this case there are Uj C j j unitary r∈ − ⊕ ≈ ∗ ∗ ∈∗ and λ C with λ = 1 such that Bj = U (A C)Uj and Vj = λjU P where P is the ∈ || ||2 j ⊕ j projection from Ckj to Cδ. abex(K) shall denote the set of absolute extreme points of K. We call A matrix exposed if there is a linear pencil = B CX of size δ such that ∗ δ×δ L − K L and ∂ L(δ) K(δ) = U AU U C unitary . In case that 0 int K one can⊆ demand D thatD is∩ monic. mexp({ K) shall| ∈ denote the matrix} exposed points∈ of K. L Of course absolute extreme points are matrix extreme; matrix extreme points are ordinary extreme. r ∗ Proof. Let A be matrix extreme and A = V BjVj where Bj K(kj), kj δ, Vj j=1 j ∈ ≤ ∈ kj ×δ r ∗ kj C 0 and j=1 Vj Vj = Iδ. Let Pj be the projection from C to the range of Vj and dj \{ } r ∗ P ∗ r ∗ the rank of Vj. Then A = j=1(PjVj) (PjBjPj )(Pj Vj) with I = j=1(PjVj) (Pj Vj) and d ×δ P PjVj C j is surjective. Hence we get δ kj dj = δ and the Vj have been already bi- ∈ P ≥ ∗ ≥ P r ∗ jective. We find Uj unitary such that Bj = Uj AUj. Now we have A = j=1(UjVj) AUjVj and Lemma 5.10 implies that the U V are scalar multiples of the identity (it is clear that j j P a matrix etreme point has to be irreducible; A = C D with C, D = 0 would imply A = P ∗CP + (1 P )∗D(1 P ) where P is the projection⊕ on the first6 coordinates and hence C or D have− the same− size as A). In a similar way we prove the additional claim for absolute extreme points. Remark 6.2. If A Sg(δ) is an matrix extreme point/absolute extreme/matrix exposed ∈ point of a matrix convex set K and U Cδ×δ unitary, then also U ∗AU is matrix extreme/absolute extreme/matrix exposed.∈ Therefore, if we say things like "K has only finitely many absolute extreme points", we mean that up to unitary equivalence there are only finitely many absolute extreme points of K. Example 6.3. Let K Sg be matrix convex. Then the extreme points/exposed points of K(1) are also matrix extreme/matrix⊆ exposed and mext(K)(δ) = mext(mconv(K(δ)))(δ) as well as mexp(K)(δ)= mexp(mconv(K(δ)))(δ). Definition 6.4. We call a non-empty set T Sg+1 a matrix cone if ⊆ r kj ×s ∗ Aj T (kj),Vj C = V AjVj T. ∈ ∈ ⇒ j ∈ j X=1 δ We call T directed if for all (A0, A1, ..., Ag) T (δ) and v C we have A0 0 and the ∗ ∈ ∈ equality v A0v = 0 implies already A1v = ... = Agv = 0. If T is a directed matrix cone, then Deh(T )= A Sg (I, A) T is called the dehomogenization of T . For T Sg+1 { ∈ | ∈ } ⊆ we write mcc(T ) for the smallest matrix cone containing T . Let S Sg be matrix convex. ⊆ We call Hom(S) = mcc( (Ik, A) k N, A S(k) ) the homogenization of S. In the { | ∈ ∈ } following we write 0s for the (g + 1)-tuple of s s zero matrices. × In ordinary convexity the homogenization of a convex set S (i.e. the set cone( (1, a) a S )) is a trivial-looking process and it is very easy to see how properties of {S translate| ∈ directly} to properties of the homogenization of S and vice versa. The homogenization of a matrix convex set is a more complicated process. It is not clear how one should compute the homogenization. We will see that the matrix extreme points of a matrix convex set S are in correspondence to the ordinary extreme rays of the levels of Hom(S). This fact comes very handy in order to use the theory of ordinary convexity for the analysis of matrix extreme points. However since homogenization is a difficult process, this does not mean that matrix extreme points are easy to understand or to compute. 36 Proposition 6.5. a (a) The mappings Hom : S Sg S is matrix convex T Sg+1 T is a directed matrix cone { ⊆ | } → { ⊆ | } Deh : T Sg+1 T is a directed matrix cone S Sg S is matrix convex { ⊆ | } → { ⊆ | } are inverses of each other. g g+1 (b) Let S S be matrix convex. Then Hom(S)= A S s, k N0,B S(k),V k×k ⊆ ∗ { ∈ | ∃ ∈ ∈ ∈ C invertible : A [0s V (I,B)V ] . ≈ ⊕ } (c) Let S Sg be matrix convex and s N. Then A S(δ) is matrix extreme in S if and ⊆ ∈ ∈ only if 0s (I, A) is ordinary extreme in Hom(S)(δ + s). ⊕ (d) (Effros-Winkler separation for matrix cones) Let T Sg be a directed matrix cone g+1 ⊆ ϕ : S (δ) R linear such that ϕ(T (δ) 0 ) R>0. Then there exists a linear pencil → \{ } ⊆ g+1 H X + ... + HgXg of size δ such that T 0 A S [H X + ... + HgXg](A) 0 0 \ { } ⊆ { ∈ | 0 0 ≻ 0 = A Sg+1 V Csize(A)×δ 0 : ϕ(V ∗AV ) > 0 } { ∈ | ∀ ∈ \ { } } (e) Let S Sg be matrix convex and s N. Then A S(δ) is matrix exposed in S if and ⊆ ∈ ∈ only if 0s (I, A) is exposed in Hom(S)(δ + s). ⊕ (f) Let S Sg be matrix convex and compact. Then Hom(S) is closed and contains no non-trivial⊆ linear subspaces. (g) Let T Sg+1 be a directed matrix convex cone, B T (δ) and V Cδ×δ invertible. Then B⊆is extreme/exposed if and only V ∗BV is extreme/exposed.∈ ∈ Proof. (a) Well-definedness of both maps is easy to see. Let S Sg be matrix convex. The ⊆ inclusion S Deh(Hom(S)) is trivial. So let (I, A) Hom(S)(k). Choose Aj S(kj ) k⊆j ×k ∗ ∈ ∗ ∈ and Vj C such that (I, A)= j Vj (I, Ai)Vj. We conclude j Vj Vj = I. Hence A S. ∈ ∈ P P Let T Sg+1 be a directed matrix cone. Let (A′,B′) T (δ). Then there exists ⊆ g s×s ∈ ′ ′ k,s N and B S (s), A SC positive definite such that (A ,B ) 0k (A, B). ∈ 0 ∈ ∈ ≈ ⊕ Hence (A, B) T . Let D SCs×s such that D2 = A−1. Then we have (I,DBD) T . Hence (I,DBD∈ ) Hom(Deh(∈ T )) and as Hom(Deh(T )) is a matrix cone, we get∈ (A′,B′) Hom(Deh(T∈)). This shows T Hom(Deh(T )). The other inclusion is clear. ∈ ⊆ k ×δ (b) Let A Hom(S)(δ). Then there exists r N and Dj S(kj) and Wj C j ∈ r ∗ ∈ δ ∈ ∈ such that A = j=1 Wj (I, Dj)Wj. Let P : C im(P ) be the projection onto ⊥ → r j=1 ker(Wj) P. Then we have T r ∗ ∗ A 0 PW (I, Dj )WjP ≈ δ−rk(P ) ⊕ j j X=1 r ∗ ∗ and j=1 PWj WjP is positive definite. Hence we can choose some invertible Her- 2 r ∗ ∗ mitian V of the same size such that V = j=1 PWj WjP . Now we conclude P r −1 ∗ ∗ −1 B := j=1 V PWj DjWjP V S and A [0δ−rk(P ) V (I,B)V ]. ∈ ′ ≈′ P ⊕ (c) Let A S(δ) be matrix extreme. Let (Bj,Cj) Hom(S)(δ + s) and 0s (I, A) = P′ ∈ ′ ∈ ⊕ j(Bj,Cj). Since Hom(S) is directed, we see that there are (Bj,Cj) Hom(S)(δ) ′ ′ ∗ ∗ ∈ such that (Bj,Cj) = 0s (Bj,Cj). Write (Bj,Cj) = Uj (0sj Vj (I, Aj )Vj)Uj with P (⊕δ−s )×(δ−s ) δ×δ⊕ Aj S(δ sj), Vj C j j invertible and Uj C unitary. We conclude that∈ − ∈ ∈

∗ 0 (I, A)= Uj ∗ (I, Aj) 0 Vj Uj Vj j X 37 ∗ Hence there are unitary matrices Wj and λj C such that Aj = Wj AWj and ∗ ∈ 0 Vj Uj = λjWj . Therefore (Bj,Cj) is a scalar multiple of (I, A) and sj = 0. k ×δ Let 0s (I, A) be ordinary extreme in Hom(S)(δ + s). Let Bj S(kj ) and Vj C j ⊕ ∗ ∗ ∈ ∈ surjective such that j Vj Vj = I and j Vj BjVj = Aj. Then P P0 0s (I, A)= ∗ (I,Bj) 0 Vj . ⊕ Vj j X ∗ ∗ Hence for every j there is λj > 0 such that (Vj Vj,Vj BjVj) = λj(I, A). Thus Uj := Vj ∗ is unitary and Bj = UjAUj . √λj (d) This is an easy variant of the Effros-Winkler separation technique. In this homogeneous setting we have a linear functional which can be easily translated to a linear pencil (contrary to the case in Corollary 2.11 where we had to translate the affine-linear functional 1 ϕ with ϕ the linear functional from Corollary 2.11 and translation of the constant− part was non-constructive). We sketch the proof: Extend ϕ to a C-linear functional ϕ : (Cδ×δ)g+1 C. By the Riesz representation δ×δ→ g ∗ theorem we can find matrices H ,H , ..., Hg C such that ϕ(C)= tr(Hi Ci). 0 1 ∈ i=0 It is easy to see that the Hi have to be Hermitian. Now H0X0 + ... + HgXg is the g+1 kδ δ P k required pencil. Let B S (k) and v C . Write v = α=1 eα vα with vα C . ∈ ∈k×δ ⊗ ∈ Define the matrix V = v . . . vδ C . Then 1 ∈ P ∗ ∗ v [H0X0 + ... + HgXg](B)v = ... = ϕ(V BV ) (see the proof of Corollary 2.11 for more details) (e) Let 0s (I, A) be exposed in Hom(S)(s + δ). Then it is clear that (I, A) is also ⊕ exposed in Hom(S)(δ). Let ϕ : Sg+1(δ) R such that ϕ(I, A) = 0 and ϕ(Hom(S)(δ) → ∗ ∗ δ×δ \ R(I, A)) R> . Claim: mcc(Hom(S)(δ) (V V,V AV ) V C invertible ) is ⊆ 0 \ { | ∈ } a matrix cone and has empty intersection with (V ∗V,V ∗AV ) V Cδ×δ invertible . δ×δ { δ|×δ ∈ } Indeed let (Bj,Cj) Hom(S)(δ), Vj C 0 and V C invertible such that ∗ ∗ ∈ ∗ ∈ \ { } ∈ ∗ −1 ∗ −1 (V V,V AV ) = j Vj (Bj,Cj)Vj. We infer (I, A) = j(V ) Vj (Bj,Cj)VjV . Because (I, A) is extreme, we conclude that there are λj R>0 such that (I, A) = ∗ −1 ∗ P −1 ∗ −1 ∗ P∈ −1 λj(V ) Vj (Bj,Cj)VjV λj for all j with (V ) Vj (Bj,Cj)VjV = 0. Such a j exists and the claim is shown. 6

Now with (d) we ﬁnd a pencil H0X0 + ... + HgXg of size δ such that mcc(Hom(S)(δ) ∗ ∗ δ×δ g+1 \ (V V,V AV ) V C invertible ) B S [H X + ... + HgXg](B) 0 { | ∈ } ⊆ { ∈ | 0 0 ≻ } and [H0X0 + ... + HgXg](I, A) ⊁ 0 .

Now let A S(δ) be matrix exposed in S and L0 + LX a pencil of size δ exposing A. ∈ δ δ From Remark 5.25 we know that there is v = eα vα with vα C such that α=1 ⊗ ∈ ker(L + LX)(A) = span(v) and span(v , ..., v )= Cδ. We claim that 0 1 δ P g+1 D C ∗ ϕ : S (s + δ) R, tr(D )+ v (L X + ... + LgXg)(B)v → C∗ B 7→ 0 0 0 exposes 0s (I, A). Because T is directed, we only have to show that ⊕ g+1 ∗ ψ : S (δ) R,B v (L X + ... + LgXg)(B)v → 7→ 0 0 exposes (I, A) in Hom(S)(δ). Now let C S(δ t), V C(δ−t)×(δ−t) invertible, ∈ − ∈ δ×δ ∗ ∗ ∗ 0 U C unitary and B = U (0t V (I,C)V )U = U (I,C) 0 V U. Then ∈ ⊕ V ∗ 38 0 we have ψ(B) = v∗ I U ∗ [L X + LX](C) I 0 V U v 0. In ⊗ V ∗ 0 0 ⊗ ≥ case of ψ(B) = 0, we conclude t = 0 and C A. WLOG C = A since we can absorb ≈ δ a unitary matrix into V . From span(v1, ..., vδ ) = C we deduce that VU is a scalar multiple of the identity. (f) The ﬁrst part follows from (b), the second because Hom(S) is directed. (g) This follows from T (δ)= V ∗AV A T (δ) . { | ∈ } Lemma 6.6. Let K Sg be matrix convex and A K(δ). Then the following is equivalent: ⊆ ∈ (a) A is matrix extreme. (b) A / mconv(K(δ) U ∗AU U Cδ×δ unitary ). ∈ \ { | ∈ } Proof. (a) = (b): This is clear. ⇒ r ∗ δ×δ (b) = (a): Suppose that A = V BjVj with Bj K(1), ..., K(δ), Vj C 0 . ⇒ j=1 j ∈ ∈ \ { } We have to show that Bj A for all j 1, ..., r . To achieve that we can suppose that ≈ P ∈ { } ∗ K = mconv(A, B , ..., Br) and therefore K is compact. Set S = mconv(K(δ) U AU U 1 \{ | ∈ Cδ×δ unitary ) = K. Then we know that Hom(S) Hom(K) and (I, A) Hom(K) Hom(S) from} Proposition6 6.5 (a). Also from Proposition⊆ 6.5 we know that∈Hom(K) is\ closed and contains no non-trivial subspace. Hence the Minkowski theorem for cones tells us that Hom(K)(δ) is the convex hull of its extreme rays. As Hom(K)(δ) = Hom(S)(δ) there exists even an extreme ray C Hom(K)(δ) that is not contained in Hom(6 S)(δ). We ∈ have Hom(K)(δ) = Hom(S)(δ) V ∗(I, A)V V Cδ×δ invertible . Hence we know that ∪ { | ∈ } there is V Cδ×δ invertible such that C = V ∗(I, A)V is extreme in Hom(K)(δ) and not contained in∈ Hom(S)(δ). Proposition 6.5 tells us that (I, A) is an extreme ray of K. Hence A is matrix extreme in K and for all j we get Bj A. ≈ (b) = (a) also follows easily from the weak separation theorem Corollary X.34. ⇒ Lemma 6.7. Let K Sg be compact and matrix convex with kz(K) = δ < . Then K = mconv(mext(K))⊆. ∞ Proof. From Proposition 6.5 we know that Hom(K)(δ) is closed and contains no non-trivial linear subspaces. Hence the Minkowski theorem from the theory of ordinary convexity tells ∗ ∗ us that Hom(K)(δ) is the conic hull of its extreme rays. Let (0s V V, 0s V BV ) Hom(K)(δ) be extreme with B K and V invertible. Then Proposition⊕ 6.5 ⊕tells us that∈ B is matrix extreme in K. Hence∈ we see that δ mcc (I, A) A K(k) matrix extreme = Hom(K). ( ∈ )! k=1 [ From Proposition 6.5 (a) we conclude that K = mconv(mext(K)).

Theorem 6.8. (Free Minkowski theorem for matrix extreme points - first half) Let K Sg [Example 7.9, be compact and matrix convex. Then K = mconv(mext(K)). ⊆ Example 7.14] Proof. We apply Lemma 6.7 to show that kz(K(m)) is the matrix convex hull of its matrix extreme points. Since the matrix extreme points of kz(K(m)) are exactly the matrix extreme points of K in level at most m, this shows the claim. Lemma 6.9. [EHKM, Theorem 3.10] Let K Sg be matrix convex and A K(δ). Then ⊆ ∈ A is absolute extreme if and only if A is irreducible and for all ε N, B (Cδ×ε)g, ∈ ∈ C Sg(ε) ∈ A B K(δ + ε) B∗ C ∈ 39 implies B = 0. One can weaken the condition to ε = 1. Proof. [EHKM, Theorem 3.10] We prove the last part first: If B (Cδ×ε)g 0 and A B ∈ \ { } C Sg(ε) such that K(δ + ε), then we can choose v ε−1 such that Bv = 0 ∈ B∗ C ∈ ∈ S 6 A (Bv) and have K(δ + 1) (vB)∗ v∗Cv ∈ "= ": Let A be absolute extreme. We have already seen in the proof of Definition and Proposition 6.1 ⇒ A b that A is irreducible. Suppose E := K(δ + 1). Then we know that there is b∗ c ∈ A b A 0 unitary U such that U ∗ U = =: F for some z Rg. Fix i 1, ..., g . b∗ c 0 z ∈ ∈ { } The characteristic polynomials of Fi and Ei agree for each i 1, ..., g . We can sup- δ ∈ { } pose Ai is diagonal with entries λ1, ..., λδ. Now χ(F1)= j=1(λj X)(zi X), χ(E1)= δ δ 2 − − (λj X)(ci X) (bi)h (λj X). Hence j=1 − − − h=1 | | j6=h − Q Q δ P Q δ 2 (λj X)(ci zi)= (bi)h (λj X) − − | | − j Y=1 Xh=1 jY6=h In order that the degrees on both sides agree, both sides have to equal zero; in particular we obtain bi = 0. r ∗ " =": Suppose the right side holds. Let A = V BjVj where Bj K(kj), Vj ⇐ j=1 j ∈ ∈ kj ×δ r ∗ ∗ C 0 and V Vj = Iδ. WLOG let the Bj be irreducible (otherwise split V BjVj \{ } j=1 j P j into several summands and discard those which for which the matrix weights on the left and P ∗ ∗ ∗ r right side are zero). Define V = (V ... V ), C = Bj K. Then V is an isometry 1 r j=1 ∈ and V ∗CV = A. VV ∗ is a projection and the requirements say that 0 = (1 VV ∗)CV . L Now CV = (1 VV ∗)CV + VV ∗CV = V A. We also get V ∗C = AV ∗ by applying− the involution. We− conclude CVV ∗ = V AV ∗ = VV ∗C . This means ∗ ∗ ∗ ∗ B1V1V1 ... B1V1Vm V1V1 B1 ...V1VmBm ......  . . .  =  . . .  ∗ ∗ ∗ ∗ BmVmV ... BmVmV VmV B ...VmV Bm  1 m  1 1 m  The diagonal entries together with the fact that the Bj are irreducible and Theorem X.2 tell ∗ ∗ us that there is λj (0, 1] such that VjVj = λjI. In particular Vj is injective and kj δ. ∈ 1 1 1 1 ∗ ≤ Since CV = V A, we have Bj Vj = VjA for all j. Thus VjA Vj = Bj √λj √λj √λj √λj and Bj mconv(A). We conclude Bj A because A is matrix extreme in mconv(A) (Lemma∈ 5.10). ≈ Definition 6.10. Let K Cn be a convex set. We say that K is balanced if for all x K ⊆ ∈ and λ C with λ 1 already λx K. ∈ | |≤ ∈ The next lemma analyzed when we can dilate a matrix extreme point to another matrix extreme points. Using slightly different constructions for the proof, similar version have appeared earlier ([DK, Lemma 2.3]) and later ([EH, Section 2.2]). Lemma 6.11. Let K Sg be a compact matrix convex set. Let A K(m) be a matrix [Example 7.12] extreme point of K. Then⊆ either A is absolute extreme or it dilates∈ to a matrix extreme point A b B = b∗ c 40 where b (Cm)g 0 and c Rg. ∈ \ { } ∈ Proof. This proof is inspired by [M, Proposition 5.1]. Assume that A is not absolute extreme. Consider the compact convex set A b F := b (Cm)g c Rg : K(m + 1) ∈ ∃ ∈ b∗ c ∈ which is balanced due to I 0 A b I 0 A λb = for λ C with λ = 1. 0 λ∗ b∗ c 0 λ (λb)∗ c ∈ | | Choose an extreme point b = 0 of F . Afterwards choose an extreme point c of the set A b 6 A b E = c Rg K(m + 1) and set B = . ∈ b∗ c ∈ b∗ c r ∗ m+1 r ∗ Assume now B = V HjVj where the Vj = 0, Hj K(h) and V Vj = I. j=1 j 6 ∈ h=1 j j WLOG we can suppose H K(m + 1) (Justification: In the case that K(1) contains only P j S P one element a, it is easy to see∈ that a is the only matrix extreme/absolute extreme point of K; thus the theorem is true in this setting. In the case that some sj := m+1 size(Hj) > 0 − ′ and K(1) contains infintely many elements, we can find aj K(1) such that B Hj := sj ∈ 6≈ Hj α=1 aj because the Bi have only finitely many eigenvalues. Now we replace Hj ⊕ ′ ∗ m+1 size H s in the matrix convex combination by QjH Q where Qj : C C j 0 j is the L j j → ⊕ { } canonical projection). m+1 m m Let P : C C , (x1, ..., xm+1) (x1, ..., xm) be the projection on C . Then we can → 7→ W choose unitaries U such that U V P ∗ = j . We calculate j j j 0 r r W A = P BP ∗ = (W ∗ 0)U H U ∗ j = W ∗(PU H U ∗P ∗)W . j j j j 0 j j j j j j j X=1 X=1 Define J = j 1, ..., r Wj = 0 . Therefore we conclude for all j J that there are { ∈ { } | 6 ∗ ∗} ∗ ∗ ∈ m×m λj C 0 such that PUjHjU P = Z AZj and Wj = λjZ where the Zj C are ∈ \ { } j j j ∈ ∗ ∗ ∗ A sj unitary and j∈J λjλj = 1. We deduce that jUjHjUj j = ∗ with j := Zj 1 U U s tj U ⊕ j P Bh sh for some sj,tj. Define H = 1, ..., r J. For h H write Hh = ∗ . Now we { } \ ∈ sh th calculate ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ B = V U [ jUjHjU ] jUjVj + V U [UhHhU ]UhVh j j Uj U j Uj U h h h Xj∈J hX∈H ∗ ∗ ∗ A sj ∗ ∗ Bh sh = Vj Uj j ∗ jUjVj + Vh Uh ∗ UhVh, U sj tj U sh th Xj∈J hX∈H ∗ Wj ZjWj λjI where jUjVjP = j = = for j J. U U 0 0 0 ∈ We have written B as ∗ λj I 0 λjI wj 0 0 0 wh I = ∗ ∗ + ∗ ∗ wj yj 0 yj wh yh 0 yh Xj∈J hX∈H ∗ ∗ λj λj λj wj 0 0 = ∗ ∗ ∗ + ∗ ∗ λjwj wj wj + yj yj 0 w wh + y yh j∈J h∈H h h X X 41 ∗ λj I 0 A sj λjI wj 0 0 Bh sh 0 wh B = ∗ ∗ ∗ + ∗ ∗ ∗ wj yj sj tj 0 yj wh yh sh th 0 yh Xj∈J hX∈H ∗ ∗ ∗ λj λjA λj Awj + λj sjyj 0 0 = ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ + λjwj A + λjyj sj wj Awj + wj sjyj + yj sj wj + yj tjyj 0 qH Xj∈J hX∈H ∗ ∗ λj λjA λj sjyj 0 0 = ∗ ∗ ∗ ∗ ∗ ∗ + λjyjsj wj Awj + wj sjyj + yj sj wj + yj tjyj 0 qH j∈J h∈H X ∗ ∗ ∗ ∗ ∗ X where qh = whBhwh + whshyh + yhshwh + yhtyh. λ∗ Because F is balanced, we have yj j s F . Now the Cauchy-Schwarz inequality shows |yj | |λj | j ∈ ∗ ∗ ∗ yj λj yj λj that b = λ yjsj = ( λj yj ) sj is a convex combination of the sj j∈J j j∈J | || | |yj | |λj | |yj | |λj | λ∗ and 0 F . b is extreme and we conclude b = yj j s for all j J. The Cauchy-Schwarz P P |yj | |λj | j ∈ ∗ ∈ inequality implies λj = yj . Thus y yj = 1 and wℓ = 0 for all ℓ 1, ..., r , yh = 0 | | | | j∈J j ∈ { } for h H. We deduce ∈ P |yj | |λj | A b ∗ I 0 I 0 A sj yj λj A b ∗ = |y | |λ | = |yj | |λj | ∗ |yj | |λj | s tj  ∗ j j  0 ∗ b tj 0 ∗ j b ∗ tj yj λj ! yj λj ! yj λj  ∗  We see tj E and y yjtj = t for all j J. Hence t = tj by the choice of t. ∈ j∈J j ∈ Corollary 6.12. (AbsoluteP Minkowski theorem for sets with finite kz-number) Let K be [Example 7.11] a compact matrix convex set and kz(K) < . Then mconv(abex K)= K. If S Sg is a set with only irreducible elements and mconv(∞ S) = K, then S contains every element⊆ of abex(K) (up to unitary equivalence). Proof. We have mconv(mext(K)) = K due to the Minkowski theorem for matrix extreme points. Now let A mext(K). If A is absolute extreme, A abex K. Otherwise there exists B mext(K∈) and a non-trivial projection P such that∈ A = P BP ∗. However it is clear that∈ each matrix extreme point has size at most kz(K). This means that every matrix point dilates to an absolute extreme point. Hence mext(K) mconv(abex(K)). The second part is clear from the definition of absolute extreme points.⊆ The following corollary was obtained in [EHKM] by applying the Gleichstellensatz. We are able to conclude it from Corollary 6.12. This results in another way to prove the Gleichstellensatz (see the following proof). g ◦ Corollary 6.13. [EHKM, Theorem 1.2] Let L S . Then S = L = mconv(L, 0) has only finitely many absolute extreme points. Their∈ matrix convex hullD equals S.

Proof. By Corollary 6.12 and Corollary 3.7 we can choose A , ..., Ak abex S which are 1 ∈ pairwise not unitary equivalent and fulfill L 0 mconv(A1, ..., Ak). Thus A1, ..., Ak equals abex(S) (up to unitary conjugations) by⊕ Corollary∈ 6.12. For every j {1, ..., k we} ∈ { } have Aj S and hence Aj = 0 or Aj T := mconv(L). Let r = k if 0 is not an absolute ∈ ∈ extreme point of S, and r = k 1 as wells as WLOG Ak = 0 if 0 is an absolute extreme point of S. − Since A is absolute extreme in T , we know that A B L for a tuple B . Now 1 1 ⊕ 1 ≈ 1 A2 mconv(B1, A1) and the fact that A2 is absolute extreme implies that B1 A2 B2 ∈ r ≈ ⊕ for a tuple B2. Inductively we see that up to unitary equivalence H := j=1 Aj is a direct summand of L. We observe L = H. D D 42 L We have seen the following: If S is the polar of a free spectrahedron L and H is a monic D linear pencil such that H = L, then S has only finitely many absolute extreme points and the direct sum B of theD absoluteD extreme points that do not equal 0 is a direct summand of H and defines the same free spectrahedron as H. Hence I BX is the minimal pencil − defining L (this is the statement of the Gleichstellensatz in the version of Helton, Klep and McCullough;D note that B does not depend on the pencil defining our spectrahedron). Corollary 6.14. Let S Sg matrix convex, closed and 0 S. Then S is the polar of a free spectrahedron defined⊆ by a monic linear pencil if and onl∈y if S has only finitely many absolute extreme points and the matrix convex hull of those is again S. Remark 6.15. The above construction yields also another proof of Arvesons boundary theorem for finite-dimensional operator systems in a matrix algebra (Lemma 5.10). Indeed if L Sg(δ) is irreducible, we have seen mconv(L, 0) = mconv(abex(mconv(L, 0))) and it is clear∈ that only 0 and L can be absolute extreme in mconv(L). Hence L is absolute extreme. Together with Definition and Proposition 6.1 this proves the claim if we can show that there is no unitary U Cδ×δ λI λ C such that U ∗LU = L. If U ∗LU = L, ∈ \ { | ∈ } then the Li and U commute. By Burnsides theorem Theorem X.2 the algebra generated δ×δ by the Li equals C and so U is in its center. This means that U is a scalar multiple of the identity. Proposition 6.16. Let K Rg convex, compact. Then A K is an exposed extreme ⊆ ∈ point of K if and only if for all B K there is ϕ : Rg R linear such that ϕ(A) < ϕ(B) and ϕ(A) ϕ(C) for all C K. ∈ → ≤ ∈ 2 Remark 6.17. Let L,H Sg(δ) and v δ with L(H) 0, ker L(H) = span(v) and ∈ ∈ S dim M(H, v)L = δ.

Write v = α eα vα = α wα eα. Set w = α eα wα. Then L(H) H(L), ⊗ ⊗ ⊗ ∗ ≈ so H(L) 0 and ker H(L) = span(w). We have (v1 ... vδ) = (w1 ... wδ) and hence P P P dim M(L, w)H = δ. Proposition 6.18. a (a) Let K Sg be matrix convex. Every matrix exposed point of K is matrix extreme. ⊆ (b) Let H Sg(δ) be irreducible. Then H is exposed in (mconv(H, 0))(δ). ∈ ∗ Proof. (a) Suppose C DX is a linear pencil exposing A and A = j VjBjVj such ∗ − ∗ that I = VjV , every V is surjective and Bj K. Assume B is not unitarily j j j ∈ 1P equivalent to A. Since ( K)(δ 1) = we know that there is v Cδ Cδ such P DC−DX ∩ − ∅ ∈ ⊗ that dim M(A, v) = δ and span(v) = ker(C DX)(A)(Remark 5.25). Therefore C−DX − (I V ∗)v = 0 implies V = 0. ⊗ 1 1 (b) Denote K = (mconv(H, 0))(δ). Due to Proposition 6.16 it is enough to show that for every L K H we find h : Sg(δ) R affine linear such that H ker(h), h(L) > 0 as well as h∈(C) \ {0 for} all C K. So fix→L K H . ∈ ≥ ∈ ∈ \ { } Case 1: If L is not unitarily equivalent to H, then L = H. Since L mconv(H, 0), we D 6 D ∈ δ2 have L H. Thus L * H and we choose Z ∂ H int( L) and v such that D ⊇ D D D ∈ D ∩ D ∈ S v∗H(Z)v = 0. Then set h(C) := v∗(I CX)(Z)v for C Sg(δ) to obtain a separating affine linear function. − ∈ ∗ Case 2: Now assume that U = I is unitary and L = U HU. Since pz( H) = δ, we can δ×6δ D choose Z ∂ H and v C 0 such that dim M(Z, v)H = δ and dim ker(H)(Z) = 1 ∈ D ∈ \ { } 2 (Theorem 5.20 and Remark 5.25). We apply Remark 6.17 to get w δ such that ker(I 43 ∈ S − ∗ ZX)(H) = span(w) and dim M(H, w)I−ZX = δ. Now it is clear that w (I ZX)(L)w > 0 because w is not an invariant subspace of I U. This leads to a separating− hyperplane as in the case before. ⊗ Lemma 6.19. Let A Sg(δ) be a matrix extreme point of a matrix convex set K Sg ∈ ⊆ with 0 K. Suppose there is linear ϕ : Sg(δ) R such that ϕ(A) = 1 and ϕ(B) < 1 for all B ∈K(δ) A . Then A is matrix exposed→ in K. ∈ \ { } Proof. Since A is matrix extreme, we know for the matrix convex set S := mconv(K(δ) \ U ∗AU U Cδ×δ unitary ) that A / S. Now the result follows directly from the separation{ | theorem∈ for non-closed} sets Corollary∈ 2.17. Theorem 6.20. (Characterization of matrix exposed points) Let K Sg be a matrix convex set and A K(δ) matrix extreme. Then A is exposed in K(δ) ⊆if and only if A is matrix exposed in∈K. Proof. Let A be exposed. Then Lemma 6.19 guarantees that A is matrix exposed (We can shift K to ensure 0 K. If A = 0, we can apply Lemma 6.19. Otherwise the result follows from Example 6.3.).∈ 6 Let A be matrix exposed and B CX be a linear pencil certifying this. Find v Cδ Cδ such that ker(B CX)(A) = span(− v) and dim M(A, v) = δ. Then D ∈ v∗(⊗B − B−CX 7→ − CX)(D)v separates A from all points from K(δ). Corollary 6.21. (a) If B Sg irreducible, then B is matrix exposed in mconv(B, 0). ∈ (b) If L is a monic linear pencil, then every matrix extreme point of L is matrix exposed. D Proof. (a) B is matrix extreme in mconv(B, 0) due to Lemma 5.10. Combine now Proposition 6.18 and Theorem 6.20. (b) It is known that every ordinary spectrahedron has only exposed faces. So every matrix extreme A L(δ) is exposed in L(δ) and due to Theorem 6.20 matrix exposed. ∈ D D We obtain a free Straszewicz theorem with the same homogenization trick already used in the proof of the Minkowski theorem. Corollary 6.22. (Free Straszewicz theorem) Let S Sg be matrix convex and compact. ⊆ Then S = mconv(mexp(S)). Proof. The proof runs completely analogeously to the proof of Lemma 6.7 using the classical Straszewicz theorem instead of the classical Minkowski theorem. 6.2. Smallest defining tuples of matrix convex sets, "infinite-dimensional" absolute extreme points and the general Gleichstellensatz. Remark 6.23. In the remainder of the chapter we need the results from the chapter about completely positive maps in the appendix. We remind the reader of Lemma X.23 and Lemma X.18. Let K Sg matrix convex and be a separable Hilbert space. Then ⊆ gH K is compact if and only if their exists L h( ) such that mconv(L)= K. We have ∈B H g k×k K = B S (k) ϕ : span(I,L , ..., Lg) C : ϕ(L)= B, ϕ is completely positive . { ∈ | ∃ 1 → } in that case. We want to analyze closed matrix convex sets and will always suppose they are given in this form. We have shown that the matrix extreme or matrix exposed points of a matrix convex compact set K = mconv(L) generate the whole set. Basically by considering the direct sum of all those points, we would end up with another operator tuple whose matrix convex 44 hull is K. However matrix extreme/matrix exposed points are not necessarily the minimal choice. In the following, we want to show that a kind of "smallest" choice of generating points exists if we allow certain kinds of infinite-dimensional absolute extreme points. The g direct sum of these points will appear as a generalized direct summand of all H h( ) with mconv(H) = K. We need the concept of approximate unitary equivalence∈B andH a characterization from Hadwin and Larson. g g In our definitions of S we have excluded points of h( ) for a separable infinite- dimensional Hilbert space. The next proposition showsB thatM if weM had allowed this points (in the sense that the matrix convex hull of L was defined as g B h( ) ϕ : span(I,L , ..., Lg) ( ) : ϕ(L)= B, ϕ is completely positive and { ∈B M | ∃ 1 →B M is a separable Hilbert space ), M } a compact matrix convex set would still be determined by all its tuples of matrices. There- fore if K = mconv(L) with L an operator tuple, we can think of L as a generalized element of K. g Proposition 6.24. Let be a separable Hilbert space and L,H h( ) and be the H ∈B H S operator system generated by the Li. Then mconv(H) mconv(L, 0) if and only if there is a completely positive map ϕ : ( ) such that ϕ(⊆L)= H. S→B H Proof. Suppose mconv(H) mconv(L, 0). By taking the polar we see ⊆ ◦ ◦ L = mconv(L, 0) mconv(H) = H . D ⊆ D Now the claim follows from Corollary X.22 The other direction is obtained by applying Lemma X.18. In 1969 Arveson introduced the notion of boundary representations of operator systems in order to understand the structure of completely positive maps and complete isometries. We will see that these are a variant of absolute extreme points (possibly infinite-dimensional). Definition 6.25. Let ( ) be a concrete operator system. Then a completely positive map ϕ : B( ) whereS⊆B isH a Hilbert space is called boundary representation for S if ϕ has onlyS → oneK completelyK positive extension ϕ : C∗( ) B( ) and ϕ is an irreducible representation. S → K Lemma 6.26. [A, Proposition 2.4] Let e( ) be a concrete operatore system and ϕ : ( ) completely positive. ThenS the ⊆B followingH is equivalent: S→B H1 (a) If ψ : ( 1 2) is a completely positive map, P denotes the projection from S→BontoH ⊕Hand P ψ(A)P ∗ = ϕ(A) for all A , then ψ = ϕ ρ for some H1 ⊕H2 H1 ∈ S ⊕ completely positive ρ : ( 2) (b) There is only one completelyS→B positiveH extension of ϕ to C∗(S) and this extension is a representation (in this case we say ϕ has the unique extension property). Proof. [A, Proposition 2.4] (a) = (b): Suppose ϕ fulfills (a). We show that each ∗ ⇒ completely positive map ϕ : C ( ) ( 1) extending ϕ is a representation which also proves the uniqueness (existence isS due→ BtoH the Arveson extension theorem Theorem X.13). By Stinesprings representatione theorem we can find a Stinespring representation ψ : C∗( ) ( ) such that ϕ(A) = P ψ(A)P ∗ for all A C∗( ) (Theorem X.14). S → B H1 ⊕H2 ∈ S Let Q be the projection from 1 2 onto 2. Using (a) we know that we can write P ψ(A)P ∗ 0 H ⊕H H ψ(A)= fore all A . As ψ ist a representation, this equality is 0 Qψ(A)Q∗ ∈ S even true for all A C∗( ), andP ψP ∗ is also a representation. ∈ S 45 (b) = (a): Suppose ϕ fulfills (b) and ψ : ( 1 2) is a completely positive map and P⇒ ψ(A)P ∗ = ϕ(A) for all A . By ArvesonsS→B H extension⊕H theorem Theorem X.13 we ∈ S ∗ can extend ψ to a completely positive map ψ : C ( ) ( 1 2). From (b) we know that P ψP ∗ is a representation. For A C∗( ) we haveS →B H ⊕H ∈ S P ψ(A)∗P ∗P ψ(A)P ∗ = P ψ(A∗)P ∗P ψ(A)P ∗ = P ψ(A∗A)P ∗ P ψ(A)∗ψ(A)P ∗ by the Schwarz inequality Corollary X.26. This means 0 P ψ(A∗)[1 P ∗P ]ψ(A)P ∗ = P ψ(A∗)[1 P ∗P ]∗[1 P ∗P ]ψ(A)P ∗ which implies [1 P ∗P]ψ(A)P ∗ = 0−. Thus ψ is of the form ψ = −P ψP ∗ ρ.− − ⊕ g Corollary 6.27. Let be a separable Hilbert space, L h( ) , the operator system H g ∈B H S generated by Li and A S (δ). Then A is an absolute extreme point of mconv(L) if and ∈ only if the unital linear map ϕ : S Cδ×δ defined by ϕ(L)= A is a well-defined boundary representation. → Proof. This is true due to Lemma X.18, Lemma 6.26 and Lemma 6.9.

Consider a compact matrix convex set set K Sg and write K = mconv(L) with g ⊆ L h( ) . We see that a boundary representation of the operator system generated ∈ B H by the Li is basically a (possibly) infinite-dimensional absolute extreme point of K. These boundary representations do not depend on the choice of the generator L (cf. [DDSS, Theorem 5.1]). In order to answer the question of Arveson whether there are "enough boundary representations" for an operator system , Davidson and Kennedy showed that every pure completely positive map ϕ : ( S) where is separable can be dilated to a boundary representation ϕ acting againS→B on a separableH HilbertH space ("pure" can be interpreted as ϕ defining a matrix extreme point). In particular the matrix convex hull of all boundary representation of the operator system generated by L is again K. We will use some of the ideas of Davidson and Kennedy to prove the general Krein-Milman theorem/Gleichstellensatz at the end of this chapter, which shows how to extract from all boundary representations a smallest subset which generates K. In the following we will show that there are not necessarily enough absolute extreme points to generate S, so one needs also the "infinite-dimensional" absolute extreme points to re- cover the complete S. There are even compact matrix convex sets which do not have any absolute extreme points. Considering "infinite-dimensional" absolute extreme points might also lead to other disadvantages (see Remark 7.10).

Example 6.28. Let be a separable Hilbert space and T1, T2 B( ) be two isometries

H ∗ ∗ ∈∗ H ∗ i defining the Cuntz algebra . Set L = (T1 +T1 , i(T1 T1 ), T2 +T2 , (T2 T2 )). Then there is no finite-dimensional representationA of because− is simple. Therefore− the operator system S defined by L has also no non-trivialA finite-dimensionalA boundary representations. Hence abex(mconv(L)) = . ∅ The following result by Evert is the first example of a class of compact matrix convex sets having no absolute extreme points. Everts proof uses techniques from [WW] and a different description of mconv(L) than the one from Lemma X.18, the latter one being probably the reason why it is rather long. For a concrete example see [E, Section 5].

g Corollary 6.29. [E, Theorem 1.2] Let L h( ) be a tuple of compact operators. Suppose that L has no finite-dimensional reducing∈ B H subspace. Then mconv(L) has no absolute extreme points. 46 Proof. (Sketch) Let = C∗(L) ( ) and the non-unital -algebra generated by L. is a non-unitalA C∗-algebra⊆ of B compactH operators.B WLOG∗ of generality we can B suppose that C∗(L) = . Let ψ : ( ) be a non-trivial representation. Then ψ is also a representationH H of non-unitalA→B-algebras.K [C, Theorem 16.11 and Theorem |B ∗ 16.6] say that ψ decomposes into a direct sum of irreducible representations ψj and |B every ψj is unitarily equivalent to a subrepresentation of the identity representation on ∗ A ( ) A is compact = C (L). Therefore the ψj are infinite-dimensional and ψ is{ infinite-dimensional.∈ B H | Thus} there are no finite-dimensional boundary representations for . A Definition 6.30. Let be a separable Hilbert space and ( n)n a sequence of closed H H ∈N finite-dimensional subspaces of such that n m for n m and n = . If we H H ⊆H ≤ n∈N H H are given operators An : n n such that Am Hn = An for all n m and there exists H →H | g ≤S C N such that An C for all n N, we call A h( ) defined by A = An for ∈ || || ≤ ∈ ∈ B H |Hn all n N the limit of (An)n . ∈ ∈N g In case that S = mconv(L) S is a compact matrix convex set as well as all An are matrix extreme points of S and A⊆corresponds to a completely positive map that is a boundary representation of L (the operator system defined by the Li) we say that A is an accesible absolute boundaryS point of S. From Proposition 6.24 it is clear that mconv(A) mconv(S) in this case. We see that all absolute extreme points of S are accesible absolute⊆ boundary points. g g Lemma 6.31. Let S S be compact matrix convex and A h( ) the limit of a ⊆ ∈ B H sequence (An)n∈N of matrix extreme points of S. Then A is irreducible. ∗ g g Proof. Suppose A = W (B C)W with W unitary and B h( B) ,C h( C ) ⊕ ∈ B WH ∈ B H defined on a Hilbert space of dimension at least 1. Write W = A . For n N let WB ∈ ∗ Pn : n be the canonical projection and QB,n : B im(WBPn ) the projection H→H ∗ H → onto im(WBPn ) (QC,n is defined in the same manner). We obtain ∗ ∗ ∗ ∗ ∗ An = PnAPn = PnWBBWBPn + PnWC CWCPn ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ =[PnWBQB,n]QB,nBQB,n[QB,nWBPn ] + [PnWC QC,n]QC,nCQC,n[QC,nWC Pn ]

Since the An are matrix extreme, we conclude that there are λB,n, λC,n [0, 1] such ∗ ∗ ∗ ∈∗ ∗ that λB,n + λC,n = 1 as well as λB,nI = [PnWBQB,n][QB,nWBPn ] = PnWBWBPn and ∗ QB,nBQB,n is unitary equivalent to An in case that λB,n = 0 (and similar for C). Taking ∗ ∗ 6 ∗ ∗ ∗ ∗ v 1 of norm 1 we evaluate λB,n = v λB,nIv = v PnWBWBPn v = v WBWBv. Hence ∈ H ∗ λB,n =: λB is not depending on n. Remembering = n∈N n, we deduce WBWB = λBI. ∗ H H ∗ ∗ ∗ On the other hand we know WBWB = I because W is unitary. So λBWB = WBWBWB = ∗ S WB. Therefore λB = 1 and λC = 1, a contradiction. Lemma 6.32. Let S = mconv(L H) Sg be matrix convex and compact and A an accesible absolute boundary point of⊕ S. Then⊆ A mconv(L) or A mconv(H). ∈ ∈ Proof. We write A ( ) as the limit of a sequence (An)n∈N of matrix extreme points of S. We have mconv(∈ BLH H) = mconv(mconv(L) mconv(H)). Hence we know that ⊕ ∪ An mconv(L) or An mconv(H) for all n N. WLOG suppose that all An mconv(L). Then∈ A mconv(L). ∈ ∈ ∈ ∈ Remark 6.33. Let S Sg be a compact matrix convex set and A S(δ). By compactness ⊆ A b ∈ there exists b (Cδ)g and c Rg such that S and (b ) is maximal under ∈ ∈ b∗ c ∈ || 1 1|| 47 A b d this property. If d (Cδ)g, e Cg,f Rg such that b∗ c e S, then necessarily ∈ ∈ ∈ d∗ e∗ f ∈ (d1)1 = 0. The reason is that that for the function  

v1 v1 (b1)1 p : v span(eδ+1, eδ+2) v = 1 R, , { ∈ | || || }→ v2 7→ v2 (d1)1

(b ) scalar multiples of 1 1 are the unique maximizers of p in case that p = 0. (d1)1 6 Lemma 6.34. Let S Sg be a compact matrix convex set and A S matrix extreme. Suppose A does not dilate⊆ to an absolute extreme point of S. Then A dilates∈ to an accesible g absolute boundary point D h( ) . ∈B H A h Proof. Let δ = size(A). We want to dilate A to a matrix extreme point B = of h∗ g A h d S of size δ + gδ with the property that if h∗ g e S, then d = 0. We call such B d∗ e∗ f ∈ a desired dilation of A. Set A0,0 = A.   If Aj,k S(δ + jδ + k) is already defined for j 0, ..., g 1 and k 0, ..., δ 1 , then ∈ j,k ∈ { − } ∈ { − } δ+jδ+k g A b we set Zj,k := b (C ) S and choose an extreme point b of Zj,k ∈ | b∗ c ∈ g in such a way that the norm of (bj+1)k+1 C is maximal and that b is extreme in Zj,k. ∈ j,k g Aj b j+1,k A b After choose extreme c Yj := c C S . Set A = S. ∈ ∈ | b∗ c ∈ b∗ c ∈ If Aj,k S(δ + jδ + k) is already defined for j = g and k 0, ..., δ 1 , then we set ∈ j,k ∈ { − } δ+jδ+k g A b Zj,k := b (C ) S and choose an extreme point b of Zj,k in such ∈ | b∗ c ∈ g a way that the norm of (b1)k+1 C is maximal and that b is extreme in Zj,k. After choose ∈ j,k g Aj b 0,k+1 A b extreme c Yj := c C S . Set A = S. ∈ ∈ | b∗ c ∈ b∗ c ∈ By the first statements of the proof of Lemma 6.11 we conclude that all the Aj,k are matrix Aj,k d extreme in S and in case that S, then (d ) = (d ) = ... = (dg) = (d ) = d∗ e ∈ 1 1 2 1 1 1 2 g,δ ... = (dj)k = 0. Now set B = A . Now continue the same procedure by dilating B to a desired dilation C of S and so on. By construction the limit of the sequence of these elements of Sg constitutes the accesible absolute boundary point (use Lemma 6.26 and Lemma 6.31). We have arrived at another type of free Krein-Milman theorem (first half) but still minimality is an issue in this version. Corollary 6.35. Let S Sg be a compact matrix convex set. Then ⊆ mconv( A A accesible absolute boundary point of S )= S. { | } Proof. Combine Theorem 6.8 and Lemma 6.34. In order to make precise what we mean by a "minimal" generator of a compact matrix convex set, we need the concept of approximate unitary equivalence of representations. 48 Definition 6.36. Let , be a separable Hilbert spaces M H (a) Let be a C∗-algebra and ϕ : ( ), ψ : ( ) two linear maps. Then A A→B H A→B M ϕ and ψ are called approximately unitary equivalent (we write ϕ a ψ) if there ∼ exists a sequence (Un)n∈N of unitary maps Un : such that for all A we ∗ H → M ∈ A have limn→∞ ϕ(A)= Unψ(A)Un in the operator norm topology. g g ∗ (b) Let A h( ) ,B h( ) and the C -algebra generated by the Ai. We write ∈ B M ∈ B H A g A a B if there is a representation ϕ : h( ) with ϕ(A)= B, a representation ⊆ A→B H ρ : ( ) of such that ρ a idA and an isometry V : such that ϕ(A)= V ∗ρA→B(A)V forK all AA . In this∼ case ϕ is a direct summand ofHρ →(Proposition K X.4) and we say that up to approximate∈A unitary equivalence A is a direct summand of B. Lemma 6.37. [HL, Theorem 1 (3)] Let , be separable Hilbert spaces, be a C∗- subalgebra of ( ). Consider a unital completelyH M positive map ϕ : ( )A. Then the following is equivalent:B H A→B M

(a) There is a representation ρ : ( ) of such that ρ a idA and an isometry V : such that ϕ(A)=A→BV ∗ρ(A)VH for allA A . ∼ M→H ∈A ∗ (b) There exists a sequence (Vn : )n∈N of isometries such that Vn AVn ϕ(A) for n in the weak operator topologyM→H for all A . → →∞ ∈A Lemma 6.38. [H, Theorem 2.5, Theorem 5.1] Let , be separable Hilbert spaces, ( ) be a C∗-algebra and ϕ : ( ) be a representation.H M A⊆B H A→B M (a) ϕ a id if and only if ϕ is rank-preserving, i.e. for all A the Hilbert space ∼ A ∈ A dimension of im(ϕ(A)) is equal to the Hilbert space dimension of (im(A)). (b) There exists a representation ρ of such that ϕ ρ a id if and only if ϕ is rank- A ⊕ ∼ A nonincreasing, i.e. for all A the Hilbert space dimension of im(ϕ(A)) is at most ∈ A the Hilbert space dimension of (im(A)) g g Lemma 6.39. Let , be a separable Hilbert spaces and A h( ) ,B h( ) . If H K ∈ B ∗ H ∈ B K A a B a A, then there is a representation ρ : ( ) of the C -algebra generated ⊆ ⊆ A→B K A by the Ai such that ρ a id and ρ(A)= B. ∼ A g Proof. Suppose that A a B a A induced by representations ϕ : h( ) with ⊆ ⊆g ∗A→B K ϕ(A) = B and ψ : h( ) with ψ(B) = A, where is the C -algebra generated B→B H −1 B by the Bi. Obviously, we have ϕ = ψ . Due to Lemma 6.38 (b) we know that for all C the Hilbert space dimension of im(ϕ(C)) is at most the Hilbert space dimension of ∈A (im(C)) and on the other hand that for all D the Hilbert space dimension of im(ψ(D)) ∈B is at most the Hilbert space dimension of (im(D)). Hence ϕ is rank-preserving and unitary equivalent to the identity representation. g g Lemma 6.40. Let be a separable Hilbert space, L h( ) and A S (δ) be a H ∈ B H ∈ matrix extreme point of mconv(L). Then there are is a sequence (Sn)n∈N of isometries δ ∗ Sn : C such that limn (Sn) LSn = A in the operator norm topology. →H →∞ Proof. Since A mconv(L), we know that there is a sequence of isometries (Vn)n∈N ∈ 1 ⊆ Vn δ (∞) ∗ (∞) 2 (C , ) such that limn→∞ A V L Vn = 0 (Lemma X.18). Write Vn = Vn B H || − n ||  .  . j δ   with Vn (C , ). Let L denote the operator system generated by the Li and consider  ∈B H S δ×δ ∗ (∞) ∞ j ∗ j the completely positive map ϕn : L C ,C V C Vn = (Vn ) CVn . Set S → 7→ n j=1 r = 2gδ2 + 1. P 49 j j j j δ j δ×δ Now write each Vn = SnTn where Sn : C is an isometry and Tn C . Then we have → H ∈ ∞ j ∗ j ∗ j j ϕn(L)= (Tn) ((Sn) LSn)Tn. j X=1 The proof of the free Caratheodory theorem Lemma 1.8 applied to the m-th partial sum tells us that m r j ∗ j ∗ j j j ∗ j ∗ j j (Tn) ((Sn) LSn)Tn = (Tn,m) ((Sn,m) LSn,m)Tn,m j j X=1 X=1 j δ j δ×δ for some isometries Sn,m : C and Tn,m C with →H ∈ r ∞ (T j )∗T j = I (T j)∗(T j) (III). n,m n,m − n n j j m X=1 =X+1 Since L is bounded and we are only interested in the limit of the ϕn we can demand r+2 r+3 r+1 r+1 that 0 = Vn ,Vn , ... for all n N (we need the additional summand Vn LVn because the right side of (III) is not∈ zero; however one could get rid of it again with the free Caratheodory technique). By Bolzano-Weierstraß we can assume that the sequences j ∗ j j ((Sn) LSn)n, (Tn)n are normconvergent for all j 1, ..., r + 1 . Hence ∈ { } r+1 j ∗ j ∗ j j A = lim (Tn) [ lim (Sn) LSn] lim Tn. n→∞ n→∞ n→∞ j X=1 Now we invoke the hypothesis that A is matrix extreme to conclude that there is one j j ∗ j such that limn (Sn) LSn A already. →∞ ≈ g g Theorem 6.41. Let , be a separable Hilbert space, L h( ) and A h( ) be H K ∈B H ∈ B K an accesible absolute boundary point of mconv(L). Then A a L. ⊆ Proof. We want to show first that there is a sequence (Rn)n∈N of isometries Rn : ∗ K→H such that limn→∞(Rn) LRn A in the strong operator topology. If A is an absolute extreme point, this follows from≈ Lemma 6.40. So suppose that is infinite-dimensional. K Let An mconv(L) be a sequence of matrix extreme points of mconv(L) of increasing ∈ size such that Am size(An) = An and A the limit of (An)n . We remind the reader that |C ∈N mconv(A) = mconv( An n N ) (cf. Lemma X.18). Again by Lemma 6.40 we know { | ∈ } j j size(An) that for each n N there a sequence (Rn)j∈N of isometries Rn : C such j ∈∗ j → K that limj→∞(Rn) LRn An in the operator norm topology. By choosing j(n) N such j(n) ∗ ≈j(n) 1 ∈ that An (Rn ) LRn n we see that there is a sequence (Rn)n∈N of isometries || − || ≤ ∗ Rn : such that A = limn R LRn in the strong operator topology. K→H →∞ n Find now Qn such that (Rn Qn) is unitary. Let L denote the operator system generated by S the Li. We know that the completely positive map ϕ : L ( ) defined by ϕ(L)= A has S →B K ∗ the unique extension property. We want to show that limn→∞ QnLRn = 0 in the strong operator norm topology. Otherwise there exists x, a sequence yn and C > 0 such that ∗ ∗ ynQnLRnx C for arbitrary big n N, WLOG for all n N. We conclude that there || || ≥ ∈ ∈ ∗ RnLRn bn are completely positive maps ϕn : L ( C) such that ψn(L)= ∗ with S →B K⊕ bn cn ∗ ∗ ∗ b x = y Q LRnx. Now because CPU( L, ( C)) is compact with respect to the BW- n n n S B K⊕ topology (Lemma X.12) there is a subsequence (ψn)n∈N which converges to a completely 50 A b ∗ positive map ψ : L ( C) such that ψ(L)= with b x = 0. This contradicts S →B K⊕ b∗ c 6 the fact that ϕ has the unique extension property. δ×δ ∗ We define the unital completely positive map ψ : L C ,C limn→∞ RnLRn in the strong operator topology. Let be the not necessarilyS → closed algebra7→ generated by L. We A want to extend ψ onto . Let s N, α1, ..., αs 1, ..., g and consider B = Lα1 ...Lαs . We see that A ∈ ∈ { } ∗ lim RnLα1 ...Lαs Rn n→∞ ∗ ∗ ∗ ∗ ∗ ∗ ∗ = lim RnLα1 (QnQn + RnRn)Lα2 (QnQn + RnRn)Lα3 ...Lαs−1 (QnQn + RnRn)Lαs Rn n→∞ ∗ ∗ = lim (RnLα1 Rn)... lim (RnLαs Rn) n→∞ n→∞ in the strong operator topology (due to the fact that the product of (CnDn)n∈N of a sequence of uniform bounded operators Cn and a sequence (Dn)n∈N of operators converging 0 in the strong operator topology converges strongly to 0). Hence ψ : δ×δ ∗ A → C , limn→∞ RnLRn is even a -homomorphism. By taking limits we see that ψ : ∗ δ×δ ∗∗ ∗ C (L) C ,C limn→∞ RnLRn is a representation of C (L). Now the claim follows from→ Lemma 6.377→ and the fact that if we have two representation π, ρ on C∗-algebras and π dilates ρ, then ρ is a direct summand of π (Proposition X.4). We obtain the following corollary, which can be also interpreted as a general Krein-Milman theorem for compact matrix convex sets K that characterizes the "smallest" defining tuple g L h( ) with respect to a with mconv(L) = K. This tuple will be uniquely defined up∈ to B approximatelyH equivalent⊆ representations. Theorem 6.42. (Strong free Krein-Milman for compact matrix convex sets) g g Let be a separable Hilbert space, L h( ) and K = mconv(L). Let T S be a (up to unitaryH equivalence) dense at most∈B countableH subset of abex(mconv(L))⊆such that no two points in T are unitary equivalent. Then there exists a set P of infinite-dimensional accesible absolute boundary points of mconv(L) with the following properties:

(a) mconv( A∈T A C∈P C )= K. g⊕ ∗ (b) For all H h( ) with mconv(H)= L there is a representation ρ of C (H) such that L∈B H L ρ a id ∗ and an operator tuple B such that ρ(H)= A C B, ∼ C (H) A∈T ⊕ C∈P ⊕ in other words A∈T A C∈P C a H. g ⊕ ⊆ L L (c) If H h( ) is another tuple satisfying (a) and (b), then there is a representation ρ ∗∈B H L L of C (H) such that ρ a id ∗ such that ρ(H)= A C ∼ C (H) A∈T ⊕ C∈P In case that mconv(abex K)= K, we can choose P = . ∅L L Proof. (Sketch) (c) follows directly from Lemma 6.39

Suppose T is infinite. Write T = An n N and all the An pairwise not unitary { | ∈ } equivalent. Set δn = size(An) for n N and let L h( ) be the operator system ∈ S ⊆ B H ∗ generated by the Li. Theorem 6.41 implies that there is a representation ρ1 : C (L) g δ → ( ), a Hilbert space and an operator tuple B h( ) such that C 1 = , B H M1 1 ∈B M1 ⊕ M1 H ρ a id ∗ and ρ (L) = A B . As A is absolute extreme and A A , we have 1 ∼ C (L) 1 1 ⊕ 1 2 2 6≈ 1 A2 / mconv(A1). Lemma 6.32 says that A2 mconv(B1). Hence by Theorem 6.41 there ∈ ∗ ∈ exists a representation ρ2 : C (B1) ( 1), a Hilbert space 2 and an operator tuple g δ →B M M B h( ) such that C 2 = , ρ a id ∗ and ρ (B )= A B . 2 ∈B M2 ⊕ M2 M1 2 ∼ C (B1) 2 1 2 ⊕ 2 By continuing in the above way we construct for n N the closed subspaces n of δn ∈ g M n−1 with 0 = , n−1 = C n, Bn h( n) , B0 = L, representations M M H M ⊕ M51 ∈ B M ∗ ρn : C (Bn ) ( n ), ρn a id ∗ and ρn(Bn )= An Bn. −1 →B M −1 ∼ C (Bn−1) −1 ⊕ δn Set = n∈N C . Now for n N we can find a sequence of isometries (Un,m : K ⊆ H ∈ ∗ n−1 n−1)m∈N such that ρn(C) = limm→∞ U CUn,m in the operator norm for all M →L M n,m ∗ n δk 1n 0 C C (Bn−1). Let 1n : k=1 C , D D. For n,m N set Vn,m = : ∈ →H 7→ ∈ 0 Un,m n−1 δk L x x = k=1 C n−1 , . We define ρ : L ( ) H ⊕ M →H y 7→ Un,my S →B K L ∗ ∗ L ,x lim Vm,m...V1,mLV1,m...Vm,mx = An 7→ K 7→ K 7→ m→∞ ! n∈N K M in the strong operator topology. Since all the Vn,m are isometries, we can take limits and extend ρ to ρ : L ( ) S →B K ∗ ∗ L ,x lim Vm,m...V ,mLV1,m...Vm,mx = An 7→ K 7→ K 7→ m→∞ 1 Mn∈N In the same way as in the proof of Theorem 6.41 we can extend ρ to a representation ρ : C∗(L) ( ) keeping the same mapping rule (a direct sum of completely positive maps with→ the B uniqueK extension property has again the unique extension property). Now we use Lemma 6.37 in combination with Proposition X.4 to obtain: There is a representation ρ2 such that ρ ρ2 a idC∗(L) or in other words A∈T A a L. The same statement in the case of finite⊕T is∼ technically easier and follows in a similar⊆ fashion. L In case that mconv(abex K) = K, we are done. Indeed it is immediate that (a) holds with P = . Since we have used only K to define T and P , but not L, (b) holds as well ∅ (set H = L). So suppose mconv(abex K) = K. Choose a dense sequence (Em)m of 6 ∈N mext(K). We form P by following the upcoming algorithm: Set P = . Let m N 0 ∅ ∈ and Pm be a finite set of accesible absolute boundary points of K already defined. If Em mconv A C set Pm = Pm. So suppose the contrary is the ∈ A∈T ⊕ C∈Pm +1 case. By Lemma 6.34 Em dilates to an accesible absolute boundary point Cm defined on L L an infinite-dimensional separable Hilbert space m. Set Pm = Pm Cm . K +1 ∪ { } In the end we set P = m∈N Pm. By Lemma 6.7 we have mconv(K) = mconv(mext(K)). Therefore (a) is fulfilled. Basically, we repeat now the strategy of the start of the proof. For simplicity of notationS we suppose P is infinite and change the numeration of the Cm in such a order-preserving way such that P = Cm m N . Inductively we { | ∈ } show with the help of Theorem 6.41 that for all m N there is a representation τm : ∗ ∈ ∗ C (L) ( ),L A C and a representation σm on C (L) such → B K 7→ A∈T ⊕ C∈Pm that τm σm a idC∗ L . For the induction start and induction step use Lemma 6.32, ⊕ ∼ (L) L which implies P mconv(ρ (L)) because P / mconv(ρ(L)) and Pm mconv(σm(L)) 1 ∈ 2 1 ∈ +1 ∈ because Pm / mconv(τm(L)) for m N. +1 ∈ ∈ Now Lemma 6.38 says that all τm are ranknonincreasing and therefore also the representation on C∗(L) defined by L A C is ranknonincreasing. Again with 7→ A∈T ⊕ C∈P Lemma 6.38 L A C follows. Since we did only use K but not the a A∈T L C∈P L special form of ⊆L in our construction,⊕ (b) follows by setting H = L. L L Corollary 6.43. (General Gleichstellensatz) Let T be a closed matrix convex set with g 0 int(T ). Then there exists a separable Hilbert space and L h( ) such that ∈ g H ∈ B H T = L and for all other H h( ) with T = H we have L a H. D ∈B K 52 D ⊆ g ◦ Proof. We apply Theorem 6.42 to find a smallest generator L h( ) of T . Let K = g ⊥ ∈ B H ( i=1 ker(Li)) . Then L K := (L1 K, ..., Lg K ) is the desired tuple (cf. Corollary X.22). The restriction onto K is| only necessary| if 0| is not contained in the closure of the matrix convexT hull of the other matrix extreme points of T ◦.

7. Examples δ×δ Example 7.1. Let L S(C X )1 be a monic linear pencil defining a bounded spectra- [Proposition 3.12] ∈ h 2i hedron L. We see that g δ 1 because the Li and I have to be linear independent. D ≤ − Proposition X.7 states that every δ-positive map from Cδ×δ into a C∗-algebra is com- B pletely positive. Now we can find Lg+1, ..., Lδ2 −1 such that the spanC(L1, ..., Lδ2−1) con- δ2−1 tains no non-trivial positive semidefinite matrix. Hence the pencil H = I LjXj de- − j=1 δ2−1 fines a compact spectrahedron H in S , the operator system defined by theP L1, ..., Lδ2 −1 δ×δ D g equals C and kz( H ) δ (Proposition 3.12). We see L = H S . D ≤ D D ∩ Example 7.2. [HKM2, Section 2.1.2] Consider the set S = (X,Y ) S2 1 X4 Y 4 [Theorem 4.2] 0 . It is known that S(1) is convex, however not an ordinary{ spectrahedron∈ | − because− the polynomial} 1 X4 Y 4 is not an RZ-polynomial. Thus S is also not a free spectrahedron. int S is not even− matrix− convex as otherwise the requirements of Theorem 4.2 would be fulfilled. Example 7.3. Let S Sg be a spectrahedron and L a monic pencil of minimal size k defin- [Corollary 5.17] ⊆ g×g g ing S. Let H = h R h invertible and A S = hA = ( hi,jAj) { ∈ | ∈ ⇒ j=1 i∈{1,...,g} ∈ S be the group of automorphisms which leave S invariant. Then there exists a group rep- } k×k P ∗ resentation U : H U C U is unitary , h Uh such that I L(hX) = UhLUh for h H. → { ∈ | } 7→ − ∈ Proof. WLOG we suppose that there is no unitary U such that ULU ∗ = L. This is for k×k instance the case when span(I,L1, ..., Lg) = C . We can can reach this case by intro- ducing new variables if necessary. The Gleichstellensatz tells us that for each h H there ∗ ∈ is a unique unitary Uh such that (I L(hX)) = UhLU . We only have have to show that − h h Uh is multiplicative. 7→ g ∗ g Let f, h H and A S . We have Uhf (LX)(A)U = (LX)(hfA) = Li ∈ ∈ hf i=1 ⊗ g g g g g g (hf)i,jAj = Li hi,kfk,jAj = Li hi,k(fA)k = j=1 i=1 ⊗ k=1 j=1 i=1 ⊗ k=1 P ∗ ∗ ∗ UPh(LX)(fA)Uh = UhPUf (LX)(A)PUf UhP. By the uniqueness P of Uhf we getPUhf = UhUf . Example 7.4. Let L, H be -minimal and -irreducible monic linear pencils describing [Corollary 5.17] D D two spectrahedra S, T Sg. Then the pencil I [(LX) (HY )] is a description of minimal size of S T . ⊆ − ⊕ × g g Proof. We have S T = (S S ) (S T )= I−LX I−HY . Now the Gleichstellensatz Corollary 5.17 in connection× × with∩Theorem× 5.20D implies∩ D that L H is a direct summand of every other monic linear pencil defining the spectrahedron S ⊕ T . × If we take a monic linear pencil L of size δ and fix m N, then ( L(mn))n∈N can be 2 ∈ D interpreted as a free spectrahedron S in Sgm in a natural way defined by a pencil I HY . The following theorem shows: In this case the spectrahedron S "knows" that it constitutes− only some levels of a spectrahedron in g variables and the structure is reflected in the pencil I HY . − Example 7.5. Let r, δ N and : Sg(m) SCr×r linear. We can write an element of [Corollary 5.17] ∈ L 53→ Sg(m) as the evaluation of the g-tuple of generic (m m)-matrices × m m = Ej,j j,j + Ej,k j,k + Ek,j k,j (IV) Y Y Y Y j X=1 j,k∈{1X,...,m}j

B1,1 ... C1,m + iCm,1 . . .  . . .  with (B1,1, ..., Bm,m,C1,2,C2,1..., Cm,m−1,Cm−1,m)

C ,m iCm, ... Bm,m  1 − 1    in the same way as we associated with . By tensoring from the right side with n n- Y Y 2 × matrices we also associate (Sg(mn)) with Sgm (n) . Let g Sj = A S (0, ..., 0, A , 0, ..., 0) T . { ∈ | ∈ } j,j-th index

If Sj(mn) = T (n) for all j 1, ..., m and n |{z}N, then S := S is a free spectrahedron. ∈ { } ∈ 1 For the minimal pencil I HX deﬁning S we have that − m m I (I HX) Ej,j j,j + Ej,h j,h + Eh,j h,j −LY ≈ −  ⊗Y ⊗Y ⊗Y  j X=1 Xj 0 such that A µB⇒ S(δ). Hence∈ we ﬁnd λ 0 suchL that⊇A +LµB = λA and B = µ A. ± ∈ ≥ λ−1 Proposition 7.8. [AZ, Lemma 2] Let A, B SCk×k and suppose that A2x span x,Ax , ∈ ∈ { } B2x span x,Bx for all x Ck. Then there is a reducing subspace of dimension 1 or 2 of A ∈and B.{ } ∈ Proof. [AZ, Lemma 2] Choose an eigenvector x Ck 0 of A B. Set H = span x,Ax = span x,Bx . It is evident that H is an invariant∈ subspace\{ } of −A and B. Since A {and B}are Hermitian,{ } it is also reducing.

Example 7.9. (Free cube) (cf. [EHKM, Proposition 7.1]) Let g 2, S = L where [Theorem 6.8] ≥ D ∗ 1 0 g 2 Li = eie . Then S = A S i 1, ..., g : I A 0 . A S is absolute i ⊗ 0 1 { ∈ |∀ ∈ { } − i } ∈ − 2 extreme if and only if [A is ordinary extreme and irreducible] if and only if [Ai = I for every i 1, ..., g and A is irreducible]. S is the matrix convex hull of its absolute extreme points. ∈ { } If g = 2, then kz(S) = 2. If g = 3, then kz(S)= . ∞ Proof. Let A S and λ ( 1, 1) be an eigenvalue of A1 with eigenvector v. Then we can ∈ ∈ − 1 ∗ 1 ∗ find µ R small such that A = 2 (A1 + µvv , A2, ..., Ag)+ 2 (A1 µvv , A2, ..., Ag) S is not an∈ extreme point of S. − ∈ 2 On the other hand let A S be irreducible such that Ai = I for all i 1, ..., g . ∈ 2 ∗ ∈ { } A b 2 Ai + bibi Aibi + bici Suppose B = ∗ S. Then Bi = ∗ ∗ ∗ 2 I, in particular b c ∈ bi Ai + cibi bi bi + ci 2 ∗ ∗ I A + bib = I + bib . We conclude that b = 0 and A is absolute extreme. i i i Now let g = 2 and A be absolute extreme. The Ai have the eigenvalues 1 and 1. Hence the requirements of the Proposition 7.8 are fulfilled and A has a reducing− subspace of dimension 1 or 2. However A is irreducible, thus A S(1) S(2). Now let g 3, m N and n = 2m. We want to construct A S(n) absolute∈ extreme.∪ Chandler has≥ shown∈ in ∈ [DS, Theorem 1] that there exist 3 projections P ,Q,R : Cn Cn (the line under P means → that P is not necessarily surjective like most other projections in this paper) such that the 55 C∗-algebra generated by P ,Q,R equals Cn×n. Burnsides theorem Theorem X.2 implies that (P ,Q,R) is irreducible. We see that (1 2P , 1 2Q, 1 2R) S is irreducible and forms an absolute extreme point of S. − − − ∈ Remark 7.10. [DS, Theorem 1] used in the last proof is also valid for infinite-dimensional separable Hilbert spaces . There exist three projections which generate as a C∗-algebra already ( ). This showsH that a matrix convex S set can have an infinite-dimensional boundaryB representationH even though abex(S)= S already. g δ×δ Example 7.11. Let L S (δ) and suppose that span I,L1, ..., Lg = C and L is [Corollary 6.12] ∈ { } Dg bounded. Then L is the matrix convex hull of its absolute extreme points and A S (k) is absolute extremeD if and only if k = δ, A is irreducible and dim ker L(A)= kδ 1∈. − The absolute extreme points A of L are in bijective correspondence to the one-dimensional δ2 D δ subspaces span(v) of C given by some v = vα eα such that (v , ..., vδ ) is an α=1 ⊗ 1 orthonormal basis of Cδ with respect to some scalar product on Cδ depending on L. An absolute extreme point A corresponds to ker L(AP).

Proof. Because S := L is bounded, we know that the Li are linear independent and D δ×δ span L , ..., Lg A SC A 0 = 0 . This is a trivial intersection of cones. R{ 1 } ∩ { ∈ | } { } Therefore there is a non-trivial linear ϕ : SCδ×δ R such that ϕ( A SCδ×δ A 0 ) → { ∈ | } ⊆ [0, ) and ϕ(spanR L1, ..., Lg ) ( , 0]. We conclude (spanR L1, ..., Lg ) = ker ϕ. Since∞ the cone of positive{ semidefinite} ⊆ −∞ matrices is selfdual, we conclude{ that} there is a positive definite matrix B SCδ×δ such that ϕ(A)= tr(B∗A) for all A Cδ×δ. ∈ ∈ Proposition 3.12, Proposition X.7 and Corollary 6.12 imply that S is the matrix convex hull of the absolute extreme points which are elements of S(1), ..., S(δ). We know that k×m k m ∗ δ×δ span L , ..., Lg C = eje Dj,h Dj,h C , ϕ(Dj,h) = 0 (where { 1 }⊗ j=1 h=1 h ⊗ | ∈ k m ej C , eh C ). n o ∈ ∈ P P Let A S(k). Then L(A) = 0, whence dim ker L(A) kδ 1. Now suppose A ∂ L(k) ∈ 6 ≤ − ∈ D and k δ. Consider the map ψ : [(ker L(A))⊥]1×δ Ck ≤ → 1 δ k k ϕ v1 ... v1 1 δ . v eα, ..., v eα . . α ⊗ α ⊗ 7→  .  α=1 α=1 ! ϕ v1 ... vδ X X  k k    If dim ker L(A) < kδ 1 or k < δ, the dimension of domain of ϕ is bigger than the dimension of the codomain− and therefore ψ admits an element c = 0 in the kernel. We can 6 write c = (LX)(b) where b (Ck×1)g. Now for λ (0, ) ∈ ∈ ∞ A λb L(A) λ(LX)(b) L(A) λc L = λb∗ 0 ≈ λ(LX)(b)∗ I λc∗ I The Schur complement tells us that this is matrix is positive semidefinite if and only if L(A) λ2c∗c 0. This is fulfilled for small λ and thus A is not in the Arveson boundary. − 2 If on the other hand k = δ and dim ker L(A)= δ 1, we write L(A)= j,h∈{1,...,δ} Cj,h ∗ − ⊗ Fj,h where Fj,h = ejeh and Cj,h span (L1, ..., Lg) = ker ϕ for h = j and Cj,j I ∈⊥ δ 6 P − ∈ span (L1, ..., Lg). Let ker L(A) = v and write v = α=1 vα eα. Since L(A) 0, we ∗ δ ∗ ⊗ can scale v such that L(A) = vv = vαv Fα,β. Since B is positive definite, α,β=1 β ⊗P ρ : Cδ Cδ C, (u, v) ϕ(vu∗) is a scalar product. Scale B in such a way that ∗ × → 7→ P tr(B ) = 1. Now we see that the vectors vα are pairwise orthogonal with respect to ρ, thus δ v1, ..., vδ is an orthonormal basis of C with respect to ρ. { } 56 Consider the map ψ : [(ker L(A))⊥]1×δ Cδ, → δ δ ϕ (λ1v1 ... λδv1) . λ vα eα, ..., λδ vα eα . . 1 ⊗ ⊗ 7→  .  α=1 α=1 ! ϕ (λ v ... λ v ) X X  1 δ δ δ    δ δ δ Suppose λ C such that λ vα eα, ..., λδ vα eα ker ψ. Then for all ∈ 1 α=1 ⊗ α=1 ⊗ ∈ ∗ T T ∗ ∗ α 1, ..., δ we obtain 0 = tr(BPvαλ ) = tr(λ B vαP)= λ, B vα and hence λ = 0. This means∈ { that }ψ is injective. h i A b Now let b (Cδ×1)g and d Cg such that S. Due to the Schur complement ∈ ∈ b∗ d ∈ this implies that (LX)b ker(ψ)= 0 and therefore b = 0; thus A is absolute extreme. ∈ { } Along the lines of the previous arguments one also shows that a one-dimensional subspace 2 of Cδ coming from an orthonormal basis with respect to ρ induces an absolute extreme point of S. Example 7.12. We demonstrate how Lemma 6.11 works in the situation of Example 7.11. [Lemma 6.11] We set δ = 3. Define

2 0 0 1 0 0 0 1 0 0 i 0 − L = 0 1 0 , 0 2 0 , 1 0 0 , i 0 0 0− 0 1  0 0 1 0 0 0 −0 0 0 − −         0 0 1 0 0 i 0 0 0 0 0 0

0 0 0 , 0 0 0 , 0 0 1 , 0 i 0 .

       −  i 1 0 0 i 0 0 0 1 0 0 0 −         3 0 0 This corresponds to B = I and ϕ = tr in Example 7.11. L(e1) = 0 0 0 has a two- 0 0 0 dimensional kernel and e1 is matrix extreme in L because (1, e1) is extreme inHom( L)(1) D ∗ D e1 b (Lemma 7.7). We want to check if e is absolute extreme. If L with b = 0, 1 b c ∈ D 6 then ker L(e1) ker(LX)(b). Hence the rows of (LX)(b) have to be elements of span(e1). ⊆ 0 0 0 The possible form of (LX)(b) is µ 0 0 where µ,ν C (the trace has to be zero). ν 0 0 ∈ The desired form is   3 0 0 µ∗ 0 ν∗ 0 3 x5 x6 0 x2 0 x3 ∗  − −  e1 b 0 0 0000 L = ∗ 0 b c µ x 0 x5 0 x4  2  0 0 0000   ∗ ∗  ν x 0 x 0 x6  3 4    with some xj. Hom( L)(2) is an ordinary spectrahedron and we know that in order to make ∗ D ∗ e1 b e1 b matrix extreme in L, we have to make I , extreme in Hom( L)(2). b c D 2 b c D Using the characterization of the extreme points of an ordinary spectrahedron Lemma 7.7, e b∗ e b∗ we see that dim ker L 1 = 5 would imply that 1 is matrix extreme. So b c b c 57 ∗ ∗ set x = 0,x = 0,x + x = 3. Hence for µ,ν C with µµ + νν = 3 the choice 2 3 5 6 ∈ 3 3 3 0 0 µ∗ 0 ν∗ 000 0 0 0 ∗   e1 b 000 0 0 0 L = ∗ ∗ b c µ 0 0 µµ 0 µν   3 3  000 0 0 0   ∗ ∗  ν 0 0 µ ν 0 νν   3 3    constitutes an matrix extreme point of L. We choose as a solution µ = 3 and ν = 0 and obtain D ∗ 3 3 e1 b 1 0 0 0 0 2 0 2 i B := = , , 3 , 3 − , 02, 02, 02, 02 b c 0 0 0 1 0 i 0 2 2 as a new matrix extreme point of L. We want to check if B is extreme and argue B d∗ D similar as above. If L with d = 0, then ker L(B) ker(LX)(d). Hence the d e ∈ D 6 ⊆ rows of (LX)(d) have to be elements of span(e1 + e4). The possible form of (LX)(d) is 000000 000000 where κ C (the traces of two submatrices consisting of column κ 0 0 κ 0 0 ∈ 1, 3, 5 resp. 2, 4, 6 have to be zero). We maximize Re(κ) in order to make d an "extreme" choice according to the proof of Lemma 6.11. Clearly this is the case for κ = 3. The kernel B d∗ of L is then maximal for L(e) = 3e eT . We obtain d e 3 3 3 3 ∗ 1 0 0 0 0 0 0 2 0 0 2 i 0 B d 3 3 − C := = 0 0 0 , 0 1 0 , 0 0 , i 0 0 , d e      2   2  0 0 1 0 0 1 0 0 0 0 0 0 − − 3  3        0 0 0 0 i 2 − 2 0 0 0 , 0 0 0 , 03, 03  3   3   2 0 0 2 i 0 0 Now C is absolute  extreme. If not, the same methods of above would imply that there is f such that (LX)(f) = 0 is a 3 9 matrix with rows in span(e1 + e5 + e9). However the traces of the submatrices6 built of× row 1, 4, 7 resp. 2, 5, 8 resp. 3, 6, 9 must have trace 0, which is not possible.

Example 7.13. Example 7.11 in connection with Proposition X.7 and Proposition 3.12 [Proposition X.7] δ×δ ε×ε implies that for all δ N there is some ε N and a (δ 1)-positive map ϕδ : C C which is not completely∈ positive and hence∈ not δ-positive.− → Example 7.14. (Computing matrix extreme points of compact free spectrahedra) Let [Theorem 6.8] S = L be a compact free spectrahedron defined by a monic linear pencil L. Let = L I. D g+1 L − Then we have Hom(S)= (A0, A) S A0 0,I A0 + ... + LgAg 0 =: T . Justification: The inclusion{ " " is∈ obvious.| T is a⊗ matrix cone. We show} that it is (0,B⊆ ) (0,C) directed. Let (A0, A) = ∗ with D0 0 be an element of the right (0,C ) (D0, D) ≻ (B) (C) set. By applying a unitary conjugation we conclude that L ∗ L 0. (C ) I D0 + (D) k×k L ⊗ L Because L is compact, we know that C span(L , ..., Lg) contains no non-trivial D ⊗ 1 positive semidefinite matrix for all δ N and the Li are linearly independent. From (B) we deduce B = 0. Consequently∈ (C) = 0 and hence C = 0. This shows that T L L 58 is directed. Now let (A , A) T . WLOG let A 0. Then we have (I, A−1A A−1) T and so 0 ∈ 0 ≻ 0 0 ∈ A−1A A−1 S. This means (I, A−1A A−1) Hom(qS) andq therefore (A , A) 0 0 ∈ 0 0 ∈ 0 ∈ qHom(S).q q q Now that we have established Hom(S)= T , suppose that the extreme rays of the ordinary spectrahedron T (k) are given. If (A0, A) is such an extreme ray we can find unitary U such ∗ −1 −1 that U (A , A)U = 0s (B ,B) with B 0. Now B B B is a matrix extreme 0 ⊕ 0 0 ≻ 0 0 point of S and all matrix extreme points of size at mostq k areq obtained in this fashion.

8. Projection number and sequences of determinants of monic linear pencils/compatible RZ-polynomials

In chapter 5 we have seen how the sequence (fk)k∈N = (detk L)k∈N of determinants of a monic linear pencil L determines how the decomposition of L in simultaneously - irreducible and -minimal pencils looks like. In this chapter we want to analyze howD D pz( L) is encoded in this sequence of determinants. More generally we extend our result alsoD to sequences of RZ-polynomials which satisfy some natural compatibility assumptions and deﬁne matrix convex sets as well. If H Ck is a subspace and P : Ck H is the projection onto H, we will set P : Ck ⊆ → → Ck,x Px. We say that a projection P is minimal with respect to a property (R) if → im(P ) + im(Q) for every other projection Q satisfying (R).

Remark 8.1. Let h : R Rk×k be a diﬀerentiable function. Then for t R we have (det h)′(t)= tr(adj(h(t))h′→(t)) where adj(h(t)) denotes the adjugate matrix of∈ h(t).

Lemma 8.2. Let L be a monic linear pencil of size δ and fk = detk L( ). Suppose X A ∂ L such that fk(A) = 0. Then there exists a unique minimal projection P such ∈ D ∗ ∇ 6 that P AP ∂ L. ∈ D ∗ Proof. Let P be a projection such that P AP ∂ L. Then we can find a kernel vector δ ∗ δ∈ D ∗ δ v = α=1 eα vα of L(P AP ). Set w := α=1 eα wα := (Iδ P )v = α=1 eα ∗ ⊗ ⊗ ⊗ ⊗ P vα. Then w ker L(A). Of course this kernel vector is unique (Remark 8.1). So P ∈ P P span(w , ..., wδ) im(P ) and the first set is the range of the minimal projection. 1 ⊆ Lemma 8.3. Let L be a monic linear pencil of size δ, fk = detk L( ) and A ∂ L(k) X ∈ D such that fk(A) = 0. Then the unique minimal projection Q with Q fk(A)Q = fk(A) ∇ 6 ∇ ∇ (i.e. the projection onto im fk(A)) equals the unique projection P given by Lemma 8.2. ∇ Proof. Because fk(A) = 0 we have dim ker L(A) = 1. We know that there is a neigh- ∇ 6 g g borhood U of A such that M := B S (k) U fk(B) = 0 = B S (k) { ∈ ∩ |1 } { ∈ 2 ∩ U dim ker L(B) = 1 = ∂ L(k) U is a connected C -manifold of dimension gk 1. | } D ∩ − Let v ker L(A) 0 . We want to calculate the normal vector NA of this manifold M in A. ∈ \ { } We have 0 = L(A)v, thus (LX)(A)v = v and 1 is the biggest eigenvalue of (LX)(A). Consider the function h : Sn(k) δk, (B, w) L(B)w. Therefore the derivative of h in δk × S 7→ ⊥ direction in (A, v) is given by L(A) ⊥ and the image of this part of the derivative is v . S |v Now if H Sn(k), then the derivative in (A, v) in direction H is given by (LX)(H)v and ∈ the component in direction of v by v, (LX)(H)v . In particular we have ∂Ah(A, v) = v. This means that h′(A, v) has full rankh and there isi a neighborhood V of (A, v) such that T = (B, w) Sg(k) δk h(B, w) = 0 is a gk2 1-dimensional connected C1-manifold. { ∈ × S | } 59 − Now we know from [HV, Lemma 2.1] that U L(k) has full dimension. Since L(k) is ⊥ ∩ D D convex, the hyperplane A + N separates L(k). Thus up to multiplication by scalar { } A D multiples the normal vector NA is only vector N that can satisfy: If we take some D with N, D < 0, then around (A, v) we can find a unique C1-resolution ϕ : D⊥ h i δk 7→ span D ( B(v, 1)) such that (A + E + ϕ1(E), ϕ2(E)) T for small E (hence (a + E ×+ ϕS(E∩)) M) and ϕ(0) = (0, v) (uniqueness of ϕ comes∈ from the fact that 1 ∈ 1 L(k) is convex and M = ∂ L(k) U; uniqueness of ϕ is due to the fact that the deriva- D D ∩ 2 tive of fk does not vanish around A). Now we want to apply the Implicit function theorem to determine N with that property. We saw already that the image of the derivative of h in (A, v) in direction δk is v⊥. For S B Sn(k), the component of v of the derivative of (h, A) in direction B is v, (LX)(H)v . ∈ h i Together with the Implicit function theorem this means that the normal vector NA points into the direction H Sg(k) such that v, (LX)(G)v = 0 for all G Sg(k) perpendicular ∈ h i n ∈ ∗ to H. Thus the normal direction NA is given by r(A) where r : S (k) R,B v L(B)v. ∇ → 7→ Since the equality fk = 0 defines the manifold M, we see that span(NA) = span( fk(A)) = δ ∇ span r(A). Now if we write v = α=1 eα vα, the proof of Lemma 8.2 tells us that P is ∇ ⊗ n ∗ ∗ the projection onto v1, ..., vδ . Since for every B S (k) we have v L(P BP )v = v L(B)v, { } P ∈ we get P fk(A)P = fk(A). ∇ ∇ On the other hand we have Q r(A)Q = r(A). Hence for every B Sg(k) we have ∇ ∇ ∈ v∗(LX)(B)v = tr( r(A)B) = tr(Q r(A)QB) = tr( r(A)QBQ)= v∗(LX)(QBQ)v. ∇ ∇ ∇ ∗ ∗ Thus we conclude 0 = v L(A)v = v L(QAQ)v. Since L is matrix convex, QAQ L. D ∈ D Hence QAQ ∂ L and im(P ) im(Q). Due to P fk(A)P = fk(A), even im(P ) = ∈ D ⊆ ∇ ∇ im(Q). g Definition 8.4. We call a system of functions (fk : S (k) R)k∈N a compatible sequence of RZ-polynomials if the following holds: → g (1) For all k N the function fk is an RZ-polynomial on S (k) with fk(0) = 1. ∈g (2) For Ai S (kj) we have fk( Aj)= fk (Aj), where k = kj. ∈ j j j j (3) Every f is invariant under unitary similtaries. k L Q P (4) There is k N such that fk is irreducible for k k . 0 ∈ ≥ 0 g Proposition 8.5. Let (fk : S (k) R)k∈N be a compatible sequence of RZ-polynomials. → −1 Then the closures of the connected components of fk (R 0 ) around 0 form a closed matrix convex set S. \ { } Proof. The claim follows from Lemma 1.10.

Corollary 8.6. Let (fk)k∈N be a compatible sequence of RZ-polynomials, S the generated closed matrix convex set and m N. Suppose that for each A ∂S(k) there is a projection ∈ ∈ P of rank at most m such that fk(A)= P fk(A)P . Then pz(S) m. ∇ ∇ ≤ Proof. Let A ∂S(k). Suppose ﬁrst that fk(A) = 0. Let P be a projection of rank ∈ ∇ 6 2 at most m such that fk(A) = P fk(A)P . Consider the function ϕ : C C, (s,t) ∇ ∇ → 7→ fk(sP AP + t(A P AP )). Then ϕ is an RZ-polynomial. [HV, Theorem 3.1] says that − there exists r N, B,C SCr×r such that ϕ(s,t) = det(I + sB + tC). We know that ′ ∈ ∈ ϕ (1, 1)(1, 1) = fk(A), A = 0 from [HV] (This is true because the connected component −1 h∇ i 6 ⊥ of fk (R 0 ) around 0 is convex and the hypersurface A + ( fk(A)) isolates it. If \ { } −1 ∇ fk(A), A = 0, this would contradict 0 int f (R 0 ).). h∇ i ∈ k \ { } Now let be the pencil I + BX1 + CX2. (1, 1) has a unique kernel vector v. Apart L 60L from that ϕ′(1, 1)(0, 1) = 0. Thus the proof of Lemma 8.3 shows v∗Cv = 0. Hence fk(P AP )= ϕ(1, 0) = 0.

Now for a general A ∂S(k) choose s k such that fs is irreducible. Now approx- ∈ ≥ imate A 0 by regular points B of the real variety defined by fs (Lemma 5.7). We ⊕ ∗ find a projection PB of rank at most m such that frk(PB)(PBBPB) = 0. By taking the limit we find a projection P of rank at most m such that f (P ∗(A 0)P ) = rk(P ) ⊕ 0. So there exists C ∂S(m) mconv(A, 0)(m). However C mconv(A, 0)(m) = mconv( P AP ∗ P projection∈ of rank∩ at most m 0 ) (Proposition∈ 3.4). Moreover 0 int(S),{ so not all| P ∗AP can be in int(S). } ∪ { } ∈

We do not know whether the converse of the previous corollary is also true.

9. Sequence of the degrees of the determinants of a monic linear pencil Remark 9.1. For the purposes of this chapter we need the concept of (noncommutative) rational functions. We will only briefly discuss the properties we need and refer the reader for instance to [R] or [KVV] for a more detailed exposition. A rational expression is a senseful combination consisting of noncommutative polynomials C X , functions +, ,−1 and brackets. We distinguish rational expressions from the h i · function they represent, so X1 +X2 and X2 +X1 are different rational expressions. If f is a rational expression, then we can evaluate it in a tuple of equally sized square matrices. The domain dom(f) of f consists of all tuples where all matrix inverses exist when following the arithmetic operations in f. On each level k the domain domk(f) of f is a Zariski-open set. We call two rational expressions t,s with non-empty domains equivalent if the intersection of their domains dom(t) dom(s) is non-empty and for all A dom(t) dom(s) we have t(A) = s(A). An equivalence∩ class of a rational expression with∈ non-emp∩ ty domain will be a rational function. Its domain will be the union of all the domains of the rational expressions it represents. In contrast to noncommutative polynomials, rational expressions can have a strange be- haviour. For example there exist non-trivial rational expressions which represent the zero function. Also a noncommutative expression can vanish on level k and be constantly the identity matrix on level k 1. − Lemma 9.2. [KVV, Proposition 2.1] If f is a rational expression which does not represent the zero function, then det f is not the zero function on dom(f). In this case dom(f −1)(n) is dense in (Cn×n)g for sufficiently big n. Lemma 9.3. (J. Volˇciˇc, private communication) Let f be a rational expression, k N k×k g k×k ∈g and domk(f) = . Then A (SC ) A domk(f) is a dense subset of (SC ) in the Euclidean6 topology∅ and{ Zariski∈ open.| ∈ }

Proof. Let be the tuple of generic matrices of size k. Since domk(f) is Zariski-open, it X is enough to show the following. Let g C[ ] be a polynomial vanishing on all tuples of Hermitian matrices, then g vanishes also∈ onX all tuples of complex matrices. However such a g vanishes for all substitutions of the i by real numbers. Therefore g vanishes also Xα,β for all substitutions of the i by complex numbers and hence g = 0. Xα,β g Proposition 9.4. [NT1, Corollary 2.3] Let L S (δ). Then the degree of p = detk L( ) ∈ X equals max rk(LX)(B) B (SCk×k)g . { | ∈ } 61 1 g Proof. We consider the polynomial q = det(Iδk Z L1 ... Lg ) i ⊗ − ⊗ Xδk − j − ⊗ X ∈ C[Z, 1 i g, 1 α, β k]. Then we write q = Z hj where Z does Xα,β | ≤ ≤ ≤ ≤ j=0 not appear in h . We know that deg(p)= α if and only if Z has multiplicity exactly δk α j P as a factor of q. −

Now suppose Z has multiplicity δk α as a factor in in q. Then hδk−α = 0. Assume that − i 6 hδk−α equals 0 for all substitutions of the α,β by real numbers. Then hδk−α equals 0 for i X all substitutions of the by complex numbers and thus hδk α=0. Therefore we know Xα,β − that there exists B (SCk×k)g such that (LX)(B) has kernel of dimension exacly δk α. So the rank is α. ∈ − k×k g On the other hand h0 = ... = hδk−α−1 = 0 and for all B (SC ) the kernel of (LX)(B) has dimension at least δk α, so the rank is at most α.∈ − Remark 9.5. (WDW ∗-decomposition) [HKM2, Section 2.6.2] Let A S(C X )δ×δ be a Hermitian matrix polynomial. Then we can find a block-diagonal matrix∈D whichh i consists of blocks of the form 0 p∗ (p) or , p 0 where p is a rational function, a lower triangular matrix W with ones on the diagonal and a permutation matrix Q such that QAQT = WDW ∗. The equality is understood entry-wise as an equality of rational functions. The algorithm to calculate such a decomposition is easy to understand: Let a b∗ E∗ A = b c F ∗ E F G  with matrices E,F,G of rational functions. In the case a = 0 or c = 0 we permute A if necessary to assume a = 0. Then we have 6 6 6 1 0 0 a 0 0 1 a−1b∗ a−1E∗ A = ba−1 1 0 0 c ba−1b∗ F ∗ ba−1E∗ 0 1 0 Ea−1 0 1 0 F − Ea−1b∗ G −E(a−1E∗) 0 0 1  − − Now one can continue with  the submatrix appearing when we del ete the first row and the first column. In the case where both a, c are zero we have 0 b∗ E∗ A = b 0 F ∗ E F G  If b = 0 and E∗ = 0 we can continue by deleting the first row and column. If b = 0 and E∗ = 0, we can permute A again to assume b = 0. So let b = 0. Then 6 6 6 1 0 0 0 b∗ 0 1 0 b−1F ∗ A = 0 1 0 b 0 0 . 0 1 (b∗)−1E∗ F (b∗)−1 Eb−1 1 0 0 G F (b∗)−1E∗ Eb−1F ∗ 0 0 1  − − Corollary 9.6. Let L be a monic linear pencil. The WDW ∗-decomposition gives an algorithm to determine deg(detk L( )) for almost all k N simultaneously. X ∈ Proof. Proposition 9.4 tells us that we have to calculate the maximal rank of (LX)(B) where B (SCk×k)g. Notice that for fixed k the set of B (SCk×k)g for which (LX)(B) ∈ ∈ has maximal rank is open in (SCk×k)g in the Euclidean topology. Let WDW ∗ the decomposition of (LX) from Remark 9.5. Let d1, ..., dh be the entries of D which are not 62{ } representing the zero function and w1, ..., wr be the entries of W . Lemma 9.2 tells us { } −1 k×k g that the intersection Dk of the domains all (dβ) and wγ and (C ) is nonempty if k k×k g is big enough. By Lemma 9.3 we know that for large k the set Dk (SC ) is non- ∩ empty as well as Zariski open in (SCk×k)g and therefore contains some B (SCk×k)g for ∈ which (LX)(B) has maximal rank. Hence for those k we can identify the maximal rank of (LX)(B) where B (SCk×k)g with the maximal rank of D(B) where B (SCk×k)g. ∈ ∈ Corollary 9.7. Let L be a monic linear pencil. Then there is some b deg(det L( )) ≥ 1 X and N N such that for all n N we have deg(det L( ))n deg(detn L( )) bn and ∈ ∈ 1 X ≤ X ≤ for n N already deg(detn L( )) = bn. ≥ X Proof. The second statement for large numbers follows directly from the end of the proof of Corollary 9.6. It remains to show that deg(det L( ))n deg(detn L( )) bn. 1 X ≤ X ≤ Let 1, ..., n be g-tuples of generic 1 1-matrices in pairwise different variables. For Y Y n × n n N we have that det L( j) equals detn L j , thus n deg(det L( )) ∈ j=1 1 Y j=1 Y 1 X ≤ deg(detn L( )). Now choose M N with Mn N. With the same construction as before X Q ∈ ≥ L we see that the M-fold direct potence of detn L equals detMn L restricted to n n-block × matrices. Thus deg(detn L( ))M deg(detMnL( )) = bnM. X ≤ X Corollary 9.8. There is a monic linear pencil L and a k > 1 such that deg(detk L( )) > k deg(det L( )). X 1 X Proof. Set g = 4 and write X = a, x, y, z . We then have for { } a z 0000 0 z 0 x 000 0 0 x 0 y 00 0  LX := 0 0 y 0 z 0 0    0 0 0 z 0 x 0    0 0 0 0 x 0 y    00000 y a  −    1000000 d1 000000 1 v1 0 0 0 0 0 v1 1 00 0 00  0 d2 0 0 0 0 0  0 1 v2 0 0 0 0  0 v 1 0 0 00 0 0 d 0 0 0 0 0 0 1 v 0 0 0  2   3   3  LX =  0 0 v 1 0 00  0 0 0 d 0 0 0  00 0 1 v 0 0   3   4   4   0 0 0 v 1 0 0  0 0 0 0 d 0 0  00 0 0 1 v 0   4   5   5   0 0 0 0 v 1 0  0 0 0 0 0 d 0  000 0 01 v   5   6   6  0 0 0 0 0 v6 1  000000 d7 0000001  with v = (za−1, xz−1az−1,yx−1za−1zx−1, zy−1xz−1az−1xy−1, − − xz−1yx−1za−1zx−1yz−1, yx−1zy−1xz−1az−1xy−1zx−1) − d = ( za−1z,xz−1az−1x, yx−1za−1zx−1y,zy−1xz−1az−1xy−1z, − − xz−1yx−1za−1zx−1yz−1x, a + yx−1zy−1xz−1az−1xy−1zx−1y) − − One easily sees that d7 is zero on dom(d7)(1) and non-trivial on dom(d7)(2). Thus we conclude that for the pencil L we have deg(det1 L) = 6 and deg(det2 L) = 14.

63 X. Appendix X.1. C∗-algebras and representations.

Definition X.1. A (complex) C∗-algebra = (A, +, , 1, . , ∗) is a C-algebra (with 1), a submultiplicative norm . making (A, +A, . ) a Banach· || space|| and an involution ∗ ex- || || || || tending the complex conjugation such that for all x,y and λ C: ∈A ∈ (x∗)∗ = x, (x + y)∗ = x∗ + y∗, (yx)∗ = x∗y∗, x∗ = x and x∗x = x 2. || || || || || || || || For a Hilbert space every closed subalgebra of ( ) with the adjoint operation as involution and the operatorH norm as a norm is the standardB H example of a C∗-algebra. A representation of a C∗-algebra is a (unital) ∗-homomorphism ϕ : ( ) where is a Hilbert space. ϕ is automaticallyA continuous. If ϕ is injective,A→B thenHϕ is an Hisometry. ϕ is called irreducible if it cannot be written as a non-trivial direct sum of other representations or equivalently if ϕ( ) does not admit a non-trivial reducing subspace. For every C∗-algebra there existsA an injective representation map ϕ. Therefore is isometric to a closed subalgebraA of ( ). Even though one can write every C∗-algebraA as a concrete subalgebra of the boundedB linearH operators on a Hilbert space, sometimes it can be advantageous to work with the abstract definition. We call x selfadjoint if x = x∗ and positive if there exists y such that y = xx∗. ∈A ∈A δ×δ Theorem X.2. (Burnsides theorem) Let L1, ..., Lg C and suppose there is some i ∗ δ×δ ∈ δ×δ with Li = 0. Then the C -subalgebra of C defined by the L , ..., Lg equals C 6 A 1 if and only if the Li have no non-trivial common reducing subspace if and only if for all v, w Cδ 0 there exists A such that Av = w. ∈ \ { } ∈A Theorem X.3. Let be a finite-dimensional C∗-algebra (as a C-vector space). Then there A m δ ×δ exist δ , ..., δm N uniquely determined up to permutation such that = C j j . 1 ∈ A ∼ j=1 L ∗ ψ(A) ρ1(A) Proposition X.4. Let be a C -algebra and φ : ( 1 2), A A A→B H ⊕H 7→ ρ2(A) η(A) a representation. If ψ is a representation, then ρ = 0.

Proof. For A, B we have ∈A ψ(A)ψ(B) ρ (AB) 1 = φ(AB)= φ(A)φ(B) ρ (AB) η(AB) 2 ψ(A)ψ(B)+ ρ (A)ρ (B) ψ(A)ρ (B)+ ρ (A)η(B) = 1 2 1 1 ρ (A)ψ(B)+ η(A)ρ (B) ρ (A)ρ (B)+ η(A)η(B) 2 2 2 1 ∗ ∗ For A selfadjoint and B = A we obtain ρ2(A) = ρ1(A) ; thus ρ1(A)ρ1(A) = 0 and ρ (A) = 0, ρ (A) = 0. Since the linear span of the self-adjoint elements is , we get 1 2 A ρ1 = 0, ρ2 = 0.

Lemma X.5. [D1, Lemma III.2.1] Let = s Ckj ×kj , ϕ : be a ∗-homomorphism A j=1 A→B into a ﬁnite-dimensional C∗-algebra Cm×m. Then there exists a unitary matrix U m×m B{ ⊆1,...,sQ} s ∈ C and uniquely determined h N , r N such that for all C = Cj ∈ 0 ∈ 0 j=1 ∈A s s L ∗ Uϕ Cj U = (Ir 0) I Cj   ⊗ ⊕  h(j) ⊗  j j M=1 M=1   64   X.2. Operator systems, completely positive maps and their connection to matrix convexity. For the study of matrix convexity we need the concepts of operator systems and completely positive maps. In particular the Arveson extension theorem and Stinespring representation theorem are important ingredients. We will sketch the proofs, but omit some details and proofs of lemmas. A good exposition of this theory is given in the book of Paulsen [P].

Definition X.6. Let ( ) be a C∗-algebra. We call a2 a , a∗ = a = A⊆B H { ∈ A | ∈ A } aa∗ a ( ) the psd (positive semidefinite) elements of . For each k N we { ∈ A | ∈ B kH } k×k A ∈ can interpret ( i=1 ) as C ( ) and take the operator norm from the former set to make the latterB oneH again a C∗⊗B-algebra.H L A (concrete) operator system (S, P ) in is a tuple satisfying: There is a linear subspace A of which contains 1 and is selfadjoint, i.e. ∗ = such that S = Ck×k S A S S k∈N ⊗ S and P = A Ck×k A 0 . P is called the set of psd elements of S. In k∈N S the literature the{ inaccuracy∈ ⊗ Shas | been established} to identify with (S, P ), which can be dangerousS because P depends on the ambient space . HoweverS we will stick to this slightly dangerous convention, reminding the reader to beA careful. If is another C∗-algebra, then ϕ : is called (unital) completely positive if ϕ is B S→B k×k k×k linear, ϕ(1) = 1 and for all k N the linear map ϕk : C C defined by B A B ϕ(A) is positive,∈ i.e. it maps positive semidefinite⊗S elements → (i.⊗Be. elements of P )⊗ to positive7→ ⊗ semidefinite elements. The condition 1 guarantees that is the span of its positive semidefinite elements; thus the condition∈ toS be completely positiveS is not merely a condition speaking about some small part of . A completely positive map ϕ is S automatically continuous and its operator norm equals 1 (even all ϕk have operator norm 1). One reason why completely positive maps were introduced is that their structure is more rigid than the one of positive maps. Much more is known about completely positive maps than about positive maps. There is no general procedure to test if a map is positive while one can build a hierarchy of semidefinite programmes to test whether a map is completely positive [HKM4, Chapter 4].

Proposition X.7. [P, Theorem 3.14] Let be a C∗-algebra and ϕ : Cn×n linear. [Example 7.13] Then the following is equivalent: A → A (a) ϕ is completely positive. (b) ϕ is n-positive. n×n ∗ (c) The so-called Choi matrix A = (Aj,h)j,h∈{1,...,n} C deﬁned by Aj,h = ϕ(ejeh) is positive semideﬁnite. ∈ ⊗A

Proposition X.8. [P, Theorem 6.1] Let be a C∗-algebra, an operator system in A S and n N. Let ϕ : Cn×n be a linear map. We define the linear functional A n×n∈ S → ∗ 1 sϕ : C C by setting sϕ(eje A) = ϕ(A)j,h. The mapping ϕ sϕ is an ⊗S → h ⊗ n 7→ isomorphism between the linear maps Cn×n and the linear maps Cn×n C. The following is equivalent: S → ⊗S → (a) ϕ is completely positive. (b) ϕ is n-positive. ∗ (c) sϕ is positive and (nsϕ(ej e I)) = I. h ⊗ j,h∈{1,...,n} Proposition X.9. [P, Proposition 2.11 and Exercise 2.3] Let be an operator system S and ϕ : C be a linear functional. Then ϕ is positive if and only if ϕ is a contraction. S → 65 Lemma X.10. [P, Theorem 6.2] Let be a C∗-algebra, an operator system in and A S A n N. Then every completely positive map ϕ : Cn×n admits a completely positive ∈ S → extension ϕ : Cn×n A→ n×n Proof. From Proposition X.8 we know that sϕ : C C is positive. Hence sϕ is a ⊗S → n×n contraction (Lemma X.10) and we can extend sϕ to a linear contraction t : C C. n×n ⊗A→ t is again positive. Now we find linear ψ : C such that sψ = t. Proposition X.8 says that ψ is completely positive. A → Definition X.11. Let be an operator system and a Hilbert space. We define a S H topology on ( , ( )) by setting: A net (ϕλ)λ converges to ϕ if for all x,y and A we haveB S B H ∈ H ∈ S ϕλ(A)x,y ϕ(A)x,y h i → h i The BW-topology can be interpreted as the weak∗-topology of a certain Banach space. An important fact is that the BW-topology turns the set of completely positive maps into a compact set. The proof appearing in most textbooks shows that the BW-topology can be interpreted as the weak∗-topology coming from forming the dual of a certain Banach space (predual). After that one can apply the Banach-Alaoglu-theorem. Instead of that, we will not introduce the predual, but adapt the proof of the Banach-Alaoglu-theorem to this setting. Lemma X.12. [P, Theorem 7.4] Let be an operator system and a Hilbert space. Then S H CPU( , ( )) := ϕ : ( ) is completely positive S B H { S→B H } is a compact set with respect to the BW-topology. Proof. Consider the map Ψ : CPU( , ( )) a C a A x y , S B H → { ∈ | || || ≤ || || · || || · || ||} A∈SY, x,y∈H f [(A,x,y) f(A)x,y ] 7→ 7→ h i where the left side carries the BW-topology and the right side the product topology. Tykhonovs theorem says that the right side is compact. It is easy to see that Ψ is a homeomorphism onto Ψ(CPU( , ( ))). So it is enough to show that Ψ(CPU( , ( ))) S B H S B H is closed in a C a A x y . The set Ψ(CPU( , ( ))) equals A∈S, x,y∈H{ ∈ | || || ≤ || ||·|| ||·|| ||} S B H Q q q(A + λB, x, y)= q(A,x,y)+ λq(B,x,y) { | } A,B∈S,x,y\∈H,λ∈C q q(A∗,x,y)= q(A,y,x) ∩ { | } A∈S\,x,y∈H n q q(Ah,j,xh,xj) 0 , ∩  ≥  n∈N,A ∈S,x ∈H, (1≤h,j≤n),(A ) is psd h,j=1 h,j h \ h,j h,j∈{1,...,n}  X  which is a intersection of closed sets.   Theorem X.13. (Arveson extension theorem) (Arveson 1969) [P, Theorem 7.5] Let be a C∗-algebra, an operator system in and a separable Hilbert space. ThenA every completely positiveS map ϕ : (A ) admitsH a completely positive extension ϕ : ( ). S →B H A→B H 66 Proof. We can find an increasing sequence ( n)n of finite-dimensional subspaces of H ∈N H such that n = . For each n N let n be an orthogonal complement of n in n∈N H H ∈ T H . Let Pn be the projection from onto n. Define ϕn : ( n) ( n), A H ∗ S H H S →B H B T 7→ [P ϕ(A)P ] 0, which is still completely positive. By Lemma X.10 we can extend ϕn to ⊕ L a completely positive map ψn : ( ). Now on the sequence (ψn)n∈N converges to ϕ in the BW-topology. AdditionallyA→B weH know that theS set of completely positive maps from to ( ) is compact, hence we can assume that (ψn)n∈N converges to a completely positiveA mapB Hψ which is an extension of ϕ. The following result characterizes completely positive maps as dilations of representations (the latter ones are easily seen to be completely positive, remembering that psd elements of a C∗-algebra are squares). The proof is a generalization of the GNS-construction. Since positive maps which have range C are automatically completely positive, the GNS-state- representation theorem can be seen as a special case of the following result. Theorem X.14. (Stinespring representation theorem) (Stinespring 1955) [P, Theorem 4.1] Let be a C∗-algebra, a separable Hilbert space and ϕ : ( ) a completely positiveA map. Then there existsH a Hilbert space , an isometryA→BV H ( , ) and a representation π : ( ) such that ϕ(A) = V ∗πK(A)V for every A ∈ B.H ThisK form is called Stinespring representation.A 7→ B K ∈ A Additionally, one can achieve that π( )V = , in which case the representation is called minimal. The minimal StinespringA representationH K (V,π, ) is uniquely determined up to unitary equivalence and can be obtained by restricting πKto its reducing subspace ′ = π( )V . K A H The following results relate completely positive maps with matrix convex combinations and are well-known. Theorem X.19 and Corollary X.21 are Positivstellensätze, which translate the property of containment of two free spectrahedra (i.e. L H) into a property regarding their defining pencils (i.e. a sums-of-squares representationD ⊆ D of H with L as weight). Corollary X.15. Let n,m N, Cn×n be an operator system. The completely ∈ S ⊆ positive maps ϕ : Cm×m are exactly the maps for which there exists an r N and n×m S → r ∗ ∈ V , ..., Vr C such that ϕ(A)= V AVj for every A . 1 ∈ j=1 j ∈ S Proof. Let ϕ be completely positive.P First extend ϕ to a completely positive map defined on the whole Cn×n. Consider the minimal Stinespring representation (V,π, ) of ϕ. We K have π(Cn×n)V Cm = . Thus is at most of dimension n2m. π is unitary equivalent to a direct sum of r identityK representationsK (Lemma X.5). Thus by absorbing a unitary operator into V we get ϕ(A)= V ∗(A ... A)V for every A Cn×n and r mn. Write ∗ ∗ ∗ n×m ⊕ ⊕ r ∗ ∈ ≤ r ∗ V = (V ...V ) with Vj C . Then we have V Vj = I and ϕ(A)= V AVj. 1 r ∈ j=1 j j=1 j The reverse direction is easy. P P Corollary X.16. Let S Sg. Then mconv(S)(k) = B Sg(k) C Sg : r ⊆ { ∈ | ∃ k×∈k ∃ ∈ N, A1, ..., Ar S : C = A1 ... Ar, ϕ : span(I,C1, ..., Cg) C : ϕ(C) = B, ϕ is completely∈ positive . In⊕ other⊕ words,∃ S is matrix convex if and→ only if it is closed with respect to direct sums} and taking images of completely positive maps. Lemma X.17. [D1, Lemma II.5.2] Let ( ) be a C∗-algebra containing no other A⊆B H compact operator from ( ) than 0 and ϕ : Cm×m be a completely positive map. B H A → m Then there exists a sequence (Vn)n∈N of isometries from C to such that limn→∞ ϕ(A) ∗ H || − Vn AVn = 0 for all A . || ∈A 67 g Lemma X.18. [DDSS, Proposition 3.5] If is a separable Hilbert space and L h( ) , then H ∈B H mconv(L)= mconv P LP ∗ P : im(P ) is a projection of rank at most k (k) { | H→ } k∈N g k×k = B S (k) ϕ : span(I,L , ..., Lg) C : ϕ(L)= B, ϕ is completely positive { ∈ | ∃ 1 → }

size(A) (∞) ∗ ∗ (∞) = A (Vn)n∈N (C , ) : Vn Vn = I, lim A Vn L Vn = 0 . { | ∃ ⊆B H n→∞ || − || } mconv(L) is compact.

(∞) ∞ Proof. (cf. [DDSS, Proposition 3.5]) Set L = n∈N L ( ). Note that the unital linear map ∈ B H L (∞) (∞) ψ : span(I,L , ..., L ) span(I,L , ..., Lg) 1 g → 1 (∞) defined by ψ(Li ) = Li is completely positive (even completely isometric). Let ϕ : k×k g span(I,L1, ..., Lg) C be completely positive and ϕ(L) = B S (δ). Since the composition of completely→ positive maps is again completely positive,∈ we know that there (∞) (∞) (∞) exists a completely positive map ρ : span(I,L1 , ..., Lg ) such that ρ(L )= B. Since ∗ (∞) (∞) C (I,L1 , ..., Lg ) does not contain other compact operators than 0, Lemma X.17 ap- δ (∞) plies and we find a sequence of isometries (Vn)n∈N (C , ) such that limn→∞ B ∗ (∞) ⊆B H || − V L Vn = 0. n || δ (∞) Now suppose (Vn)n∈N (C , ) is a sequence of isometries satisfying limn→∞ A ∗ (∞) ⊆ B H δ || − Vn L Vn = 0. Fix an isometry W (C , ). Let ε > 0. Choose n N such || ∗ (∞) ε ∈ B H ∈ that A Vn L Vn < 3 . As Vn is defined on a finite-dimensional vector space, we || − || ∗ ∗ (∞) ∗ ∗ (∞) ε know that there is an M N such that Vn PmPmL PmPmVn Vn L Vn < 3 for ∈ || (∞) m− || all m > M, where Pm denotes the projection of onto . With the same ar- ∗ ∗ H H gument we know that I Vn PmPmVn 0 for m . Hence we can find a big || −∗ ∗ ||∗ → ∗ →∗ ∞ ε m > M such that I Vn PmPmVnW LW I Vn PmPmVn < 3 . All in all, we ob- ∗ ∗ m|| − ∗ ∗ ∗− ∗ || ∗ tain A (V P L PmVn + I V P PmVnW LW I V P LPmVn) < ε. || − n m p − n m p − n m || We have proven that for eachp ε > 0 there is n Npand an isometry W (Cδ, n) ∗ n ∈ n ∈ B H such that A W L W < ε. For h 1, ..., n let Qh : , (x1, ..., xn) || − || ∈ { } H → H ∗ n 7→ xh and Rh be the projection from onto im(QhW ). Then we calculate W L W = n ∗ ∗ n ∗ ∗ H∗ ∗ ∗ h=1 W QhLQhW = h=1 W QhRh(RhLRh)RhQhW mconv( P LP P : im(P ) projection of rank at most δ ) ∈ { | H → P P } Now it is clear that mconv P LP ∗ P : im(P ) projection on a finite-dimensional space) { | H→ } g k×k B S (k) ϕ : span(I,L , ..., Lg) C : ϕ(L)= B, ϕ is completely positive . ⊆{ ∈ | ∃ 1 → } We only have to show that the latter set is closed. This is the case due to Lemma X.12. It is evident that mconv(L) is bounded. The following Positivstellensätze are very important and tell us that for two monic pencils L and H we have L H if and only if H has a sums-of-squares representation with L as a weight. D ⊆ D Theorem X.19. [HKM1][Theorem 4.6] Let be a separable Hilbert space and L,H g H ∈ h( ) , n N and define the operator systems = span(I 1,L1 0, ..., Lg 0), B H ∈ 68 S ⊕ ⊕ ⊕ ∗ = span(1,H , ..., Hg). Consider the linear map ϕ : C ( ) defined by ϕ(I 1) = T 1 S → T ⊕ I, ϕ(Li 0) = Hi. Then ⊕ (a) ϕ is well-defined and n-positive if and only if L(n) H(n) D ⊆ D (b) ϕ is well-defined and completely positive if and only if L H. D ⊆ D Proof. [Z][Theorem 2.5] (a) Suppose first that ϕ is n-positive and well-defined. Now let g A S (n). If A L(n), then (I AX)(L 0) ((I 1) [L 0]X)(A) 0 and ∈ ∈ Dn×n − ⊕ ≈ ⊕ − ⊕ (I AX)(L 0) C . Hence 0 ϕn((I AX)(L 0)) = (I AX)(H) H(A). − ⊕ ∈ ⊗ S − ⊕ − ≈ n Now suppose that L(n) H(n). We show well-definedness first. Suppose λ C such g D ⊆ D g ∈ that i=1 λiLi = 0. Write λ = µ + iν with µ,ν R . We know that Rµ L. If g ∈ ⊆ D i=1 λiHi = 0, then we find r R such that rλ / H. In the same way we show that ν = 0P. 6 ∈ ∈ D P n×n n×n n×n g Let B C such that B 0. Choose A0 C and A (C ) such that ∈ ⊗ S g ∈ g ∈ B = A0 (I 1) + i=1 Ai (Li 0). In particular A0 1+ i=1 Ai 0 0, whence ⊗ ⊕ ⊗ ⊕ g ⊗ ⊗ A 0. We notice that we can demand that A S (n) (indeed if L , ..., Lg is linear 0 ∈ 1 independent, then AP Sg(n) automatically; otherwise we representP B by a tuple A for ∈ which (Li i 1, ..., g , Ai = 0) is linear independent). | ∈ { } 6 ∗ Now let ε> 0. We find an invertible square matrix C such that C (A0 + εI)C = I. From ∗ g g ∗ (I C) [I (A0 +εI)+ i=1 Li Ai](I C) 0 we conclude that I I+ i=1 Li C AiC ⊗ ⊗∗ ⊗ ⊗ ∗ ⊗ ⊗ g 0. Thus C AC L(n) and by assumption C AC H(n). This means I I + i=1 Hi ∗ ∈ D P g ∈ D P ⊗g ⊗ C AiC 0 and consequently (A0+εI) I+ i=1 Ai Hi I (A0+εI)+ i=1 Hi Ai 0. ⊗ ⊗g ≈ ⊗ P⊗ By letting ε go to 0 we obtain ϕn(B)= A0 I + Ai Hi 0 as well. P⊗ i=1 ⊗ P (b) follows from (a). P Proposition X.20. [HKM4, Lemma 3.6] Suppose is a separable Hilbert space, L g H ∈ h( ) and that L(1) is bounded. Then 0 mconv(L). B H D ∈ Proof. [HKM4, Lemma 3.6] Define the operator systems = span(I,L , ..., Lg) and = S 1 T span(1). We will show that the map ϕ : defined by ϕ(Li) = 0 and ϕ(I) = 1 is S →T well-defined and completely positive. Since L is bounded, it is clear that that Li i D {k×k | ∈ 1, ..., g I is linear independent. Hence ϕ is well-defined. So let A0 C and { }}k× ∪k {g } g ∈ g A (C ) such that (A0 I) + i=1 Ai Li 0. We notice that A S and ∈ k×k ⊗ ⊗ ∈ A0 SC because of the linear independence of I,L1, ..., Lg. Assume that A0 was not ∈ k−P1 ∗ ∗ positive semidefinite. Choose v with v A0v < 0. Then 0 (v I)[(A0 I)+ g ∗ ∈ S g ∗ g ⊗ ∗ ⊗ i=1 Ai Li](v I) I v A0v+ i=1 Li v Aiv. This means that i=1 Li v Aiv 0; ⊗ ∗⊗ ≈ ⊗ ⊗ ⊗ ≻ therefore R≤0v Av L(1), which contradicts the boundedness. P ∈ D P P g Corollary X.21. [HKM4][Theorem 3.5] Let L,H S such that L(1) is bounded. Define ∈ D the operator systems = span(I,L1, ..., Lg), = span(1,H1, ..., Hg). Then L H if S ∗ T D ⊆ D and only if the linear map ϕ : C ( ) defined by ϕ(I) = I, ϕ(Li)= Hi is completely S → T r ∗ positive if and only if there is r N and matrices V , ..., Vr such that V LVj = H ∈ 1 j=1 j and r V ∗V = I. j=1 j j P Next,P we want to prove the bipolar theorem. In order to do that we determine the polar of a free spectrahedron. Corollary X.22. [HKM1, Theorem 4.6, Proposition 4.9] Let be a separable Hilbert g ◦ H ◦ space, L h( ) . Then we have L = mconv(L, 0) and mconv(L) = L. If L is even bounded,∈B thenHmconv(L, 0) = mconv(D L). D D ◦ Proof. L = mconv(L, 0) can be obtained by looking into Theorem X.19. D 69 Now let A L and B mconv(L)(δ). We know that there is a completely positive ∈ D ∈ δ×δ map ϕ : span(I,L , ..., Lg) C such that B = ϕ(L). Then we have that 0 L(A) 1 → ≈ (I AX)(L) and 0 ϕ ((I AX)(L)) = (I AX)(B). − size(A) − − Now let H mconv(L)◦. Then for every projection P : im(P ) on a finite-dimensional ∈ ∗ ∗ H→ subspace we have 0 (I HX)(P LP ) (I P LP X)(H) and thus H L. − ≈ − ∈ D The last part is a consequence of Proposition X.20. g g Lemma X.23. Let S S be matrix convex. Then S = mconv(L) for some L h( ) if and only if S is compact.⊆ ∈B H

Proof. First claim: Let S be compact. Let An n N be a dense subset of S. Then we { | ∈ } g claim mconv( An) = mconv( An n N )= S. An is an element of h( ) n∈N { | ∈ } n∈N B H where = Csize(An) due to compactness of S. We use Lemma X.18 and have only H Ln∈N L to show that for each isometry V : Ck we have V ∗ A V S. However L n∈N n m size(An) → H ∈∗ letting Pm : n=1 C be the canonical projection, we have V n∈N AnV = ∗ H∗ → ∗ ∗ ∗ m L limm V P [Pm AnP ]PmV = limm V P [ An]PmV S. →∞ m Ln∈N m →∞ m n=1 ∈ L ◦ Proof. (of TheoremL 2.3) Let S Sg be matrix convex.L Obviously S◦ = mconv(S 0 ) and hence we can assume that⊆S is closed and contains 0. Suppose first that 0 ∪ {int}S. ◦◦ ◦◦ ∈ We have to show S = S. It is clear that S S . We have S = n∈N Sn where ⊆ g Sn := S BSg (0,n) is matrix convex. We know that Sn = mconv(Ln) for some Ln h( ) ∩ ◦ S ∈B H due to Lemma X.23. We obtain S =: T =: Tn where (Tn)n∈ = is a n∈N N DI−LnX descending sequence of matrix convex compact sets (Remark 2.5). T ◦ ◦ ◦ We claim T = n∈N Tn . This proves the theorem as n∈N Tn = n∈N mconv(Ln) = S ◦ ◦ ◦ (Corollary X.22). It is immediate that Tn T for all n N. Now assume A T (δ) S◦ ◦ ⊆ S ∈ S ∈ and A / n∈N Tn . As A / n∈N Tn = S and S is closed, we find λ (0, 1) such that ∈◦ ∈ ∈ λA T S. Then for each n N we find Bn Tn(δ) such that (I λAX)(Bn) 0. Since ∈ \S S∈ ∈ − (Tn)n∈N is a descending sequence of compact sets, we can suppose that B = limn→∞ Bn exists. Let C S. Then (I BnX)(C) 0 for all n N with C B g (0,N). Hence ∈ − ∈ ∈ S B T = S◦. Obviously, (I λAX)(B) ⊁ 0. Together with λA T ◦, this means ∈ − 1 ∈ ker(I λAX)(B) = 0 . But now (I AX)(B) ⊁ (1 λ )I, which contradicts A T ◦,B − T . 6 { } − − ∈ ∈ In the case that 0 / int S, we can find an subspace U of Rg containing S(1) such that 0 ∈ is in the relative interior of S(1) with respect to U. WLOG U = Rr 0 g−r with some × { } r < g. Lemma 1.7 ensures that S Sr 0 g−r. Write S = S′ 0 g−r. Denote by the ⊆ × { } × { } polar in Sr. Now we have S◦◦ = (S′ Sg−r)◦ = S′ 0 g−r = S′ 0 g−r = S. × × { } × { } Proof. (of Lemma 2.1 (a)) Using Theorem 2.3 and Proposition 2.6 we get H S◦(δ) such ∈ that H(Y ) 0. g Corollary X.24. Let S S be closed and matrix convex. Then S = L for some g ⊆ D L h( ) if and only if 0 int(S). ∈B H ∈ Proof. Suppose 0 int S. Then S◦ is compact (Remark 2.5) and hence we find L g ∈ ◦ ◦◦ ∈ h( ) such that S = mconv(L). Now we have S = S = L (Theorem 2.3 and LemmaB H X.23). D A B Proposition X.25. (Schur complement) Let D = SCk×k and A 0. Then B∗ C ∈ ≻ D is positive semidefinite if and only if C B∗A−1B 0. −70 A B I 0 A 0 I A−1B Proof. = B∗ C B∗A−1 I 0 C B∗A−1B 0 I − Corollary X.26. (Schwarz inequality for completely positve maps) [P, Proposition 3.3] Let , be C∗-algebras and ϕ : completely positive. For A we have ϕ(A∗A)ϕB(A) ϕ(A∗A). A →B ∈ A 1 1 A Proof. [P, Proposition 3.3] D = 1 A = is positive semidefinite. A∗ A∗ A∗A 1 ϕ(A) Therefore 0 ϕ (D)= . Now the claim follows from Proposition X.25. 2 ϕ(A∗) ϕ(A∗A)

X.3. Real closed fields and semialgebraic sets. a A real closed field R is a field such that for R2 = r2 r R we have R2 R2 = R, R2 R2 = 0 , R2 + R2 R2 and every polynomial{ | f∈ }R[X] of odd degree∪− has a root∩− in R. The{ } convention a⊆ b : a b R2 defines an∈ ordering on R. This is the only ordering on R. One can≥ also⇐⇒ characterize− ∈ real closed fields as orderable fields R for

which R[i] is algebraically closed. Every ordered field (K, ) admits a real closure (R, ) which is a real closed extension field (R, ) extending the≤ order of K such that R K≤is algebraic. Up to order isomorphism the real≤ closure is uniquely determined. Of course| the most important example of a real closed field is the field of real numbers R. Definition X.27. Let R be a real closed field and K R a real closed subfield. We n ⊆ say that S R is a K-semialgebraic set if there exist fh, gh,j K[X] such that S = r ⊆n ∈ x R fh(x) = 0, gh, (x) > 0, ..., gh,m(x) > 0 . If we dont specify K, we mean h=1{ ∈ | 1 } K = R or K = R. S n n 2 n The sets x R i=1 xi ε where ε R>0 form the basis of a topology on R . One key fact{ ∈ is that| the theory≤ of real} closed fields∈ in the language of ordered rings admits P quantifier elimination (in particular if R is a real closed field, F1, F2 are real closed extension fields and ϕ(x) a formula in the language of real closed fields with parameters from R in the free variables x, then ϕ(x) is satisfiable over F1 if and only if it is satisfiable over F2). This means that projections of K-semialgebraic sets are again K-semialgebraic. An important corollary is the following transfer-principle. Theorem X.28. (Tarski transfer principle) Let R be a real closed field and F R be a r n ⊇ real closed extension field of R and n N . Let S = x R fh(x) = 0, gh, (x) > ∈ 0 h=1{ ∈ | 1 0, ..., gh,m(x) > 0 be an R-semialgebraic set where fh, gh,j R[X]. Then there is only one } n n S ∈ R-semialgebraic set SF F with SF R = S which is called the transfer of S into F . r ⊆ n ∩ We have SF = x F fh(x) = 0, gh, (x) > 0, ..., gh,m(x) > 0 . h=1{ ∈ | 1 } The importanceS of this theorem lies in the fact that one can easily generalize statements, which can be stated using semialgebraic sets, from the real numbers to all real closed extension fields. As an exercise the reader should try to prove: Let f : R Rn be a continuous/bijective/monotoneous/... function whose graph is semialgebraic and→ R a real n closed extension field of R. Then f extends uniquely to a function fR : R R whose → graph is R-semialgebraic. fR is continuous/bijective/monotoneous/... If S is an R-semialgebraic set, then

SF real closed extension field of F Y R is called a semialgebraic class. 71 1 1 For a real closed extension field R of R denote by mR = a R N N : 0. Every Archimedean ordered field can be interpreted∈ as a subfi∈eld ± of R. Therefore every real closed extension field of R is non-Archimedean and contains infinitesimal and infinite elements. A transcendent extension L of an ordered field (K, ) admits an ordering extending . Forming the real closure, one sees that there are many≤ non-Archimedean fields. For a≤ real closed field R and a R we define a = max(a, a). ∈ | | − For a, b R we write a b if there exist N, M N such that a < N b < M a . The residue classes∈ a of this equivalence∼ class are called∈ Archimedean classes.| | T| he| residue| | class map ν is called the canonical valuation on R. It fulfills ν(a + b) min a, b , where a b ≥ { } ≥ if there is N N such that a b N. ∈e | | ≤ | | m e e The canonical residue map R R/mR, a a R is a ring homomorphisme and fore each O → O 7→m a R there is a unique b R such that b a R . This b is called the standard part of a ∈ O ∈ ∈ and denoted by st(a). st : R is a homomorphism of ordered rings meaning that for a, b the inequality a Ob implies → st(a) st(b). ∈ O ≤ ≤ Theorem X.29. (Finiteness theorem for semialgebraic classes) Let I be an index set, R a real closed field and for every j J let Sj be a semialgebraic set. Suppose that j ∈ S = for every real closed extension field of R. Then there exists a finite set j∈J R ∅ R I J such that Sj = . T⊆ j∈I ∅ X.4. Weak separationT and a proof of the free Minkowski theorem using pencils. In the theory of ordinary convexity sometimes strong separation is not possible. For example, a non-exposed point A of a convex set S Rn can not be separated strictly from S A by an affine linear function. However it⊆ is possible to separate these two sets \ { } strongly by a finite hierarchy of affine-linear functions ϕ1, ..., ϕr in the sense that ϕ1(A)= ... = ϕr(A) = 1 and for each B S there exists j 1, ..., r such that ϕ (B) = ... = ∈ ∈ { } 1 ϕj−1(B) = 1 and ϕj(B) < 1. Equivalently, by working with an infinitesimal number ε from r j a real closed extension field R of R, one can form the affine-linear function ϕ = j=1 ε ϕj, which separates the two sets strongly. The second view point was introduced in [NT2]. P When going through the idea of the Effros-Winkler translation process, it is not clear how to translate all the ϕj into pencils (since there is no concave function Ψj in Proposition 2.9 for each ϕj). However translating ϕ can be achieved with some technical difficulties. The aim of this section is to prove the free Minkowski theorem again by using pencils and not the homogenization trick from the proof in Chapter 6. The proof is much more involved, however is has also beautiful aspects in the opinion of the author. Normally, the classical Minkowski theorem is proven by an induction on the dimension and working with the faces of a convex set. However the concept of a "free face" is cumbersome because those sets would not be matrix convex anymore. The proof of Lemma X.36 shows how those "free faces" could look like and how to argue with those non-matrix convex objects. Definition X.30. Let R be a real closed field extension of R with algebraic closure C = g r ∗ R[i] and S S . Then we set mconvR(S) := j=1 Vj BjVj r, k N,Bj S(kj),Vj k ×k r ⊆ ∗ { | ∈ ∈ ∈ C j , V BjVj = I . j=1 j } P PropositionP X.31. Let H Sg be matrix convex, 0 H and A Sg(δ) H. Set 2 ⊆ ∈ ∈ \ r = δ g 1. If R is a real closed extension field R, C = R[i] and ε R>0 infinitesimal, − g ∈ r j then there exist ϕ , ..., ϕr : S (δ) R affine linear such that ϕ := ε ϕj fulfills: 0 → j=0 ϕ(B) γ := 1+ ε + ... + εr εr+1 for all B mconv(H) and ϕ(A)= γ + εr+1. P ≤ − 72∈ δ×δ g Now extend ϕ to a R-linear function on (SC ) . For all A1, ..., As H and W1, ..., Ws size(A )×δ s ∗ s ∗ ∈ ∈ R h such that W Wh I and C = W AhWh we have ϕ(C) γ. h=1 h h=1 h ≤ −1 Proof. We construct theP ϕj inductively. For thisP purpose we set Hj = α 1. For B HR(k) := (mconvR H)(k) and a contraction ∈ δ×k δ×δ ∈ ∗ ∗ V C define fB,V : SC R, T tr(V TV ) ψ(V BV ). ∈ → 7→ − δ×k ∗ Then the set = fB,V k N, B HR(k),V C ,V V I is R-convex. For every F { | ∈ ∈ ∈ } f there is a T R,δ such that f(T ) 0. ∈ F ∈T ≥ Proof. This can be obtained by applying the Tarski-transfer to Proposition 2.9. Lemma X.33. Suppose we are in the situation of Proposition X.32. Then there is a real closed extension field of R and T ,δ such that f(T ) 0 for all f . R ∈TR ≥ ∈ F Proof. For f consider the R-semialgebraic classes ∈ F ( , T ) real closed extension field of R, T ,δ,f(T ) 0 . { R | R ∈TR ≥ } Theorem X.29 tells us that in order to prove

( , T ) real closed extension field of R, T ,δ,f(T ) 0 = , { R | R ∈TR ≥ } 6 ∅ f\∈F it is enough to show that every finite intersection of those sets is non-empty. That the latter is the case follows from Lemma 2.10 and the Tarski-transfer principle. (We remark that we cannot use the Tarski-transfer principle directly and avoid the introduction of a new real closed field; the reason is that H does not need to be R-semialgebraic. In our argument it is enough to consider finite subsets of H which are R-semialgebraic.) Corollary X.34. (Effros-Winkler - weak separation) Suppose we are in the situation of j j δ×δ 2 Proposition X.32. Then there exist L0, ..., Lg SC for j 0, ..., (g + 1)δ 1 such (g+1)δ2−1 j j j j ∈ ∈ { − } that L = ε (L + L X + ... + LgXg) fulfills: For all B H: L(B) 0 and j=0 0 1 1 ∈ R L(A) 0. R P Proof. We apply Proposition X.32 and Lemma X.33 with defined like in Proposition X.32 F . We conclude that there is a real closed extension field of R and T R,δ such that f(T ) 0 for all f . Now one follows the proof of [HMR , Proposition∈ 6.4] T to see that ≥ ∈ F δ×δ there is a linear pencil L = L + L X + ... + LgXg S( ) such that L(B) 0 0 1 1 ∈ C 1 R (even L(B) R 0, but maybe L(B) R 0) for all B H and L(A) R 0. 73 ∈ j j δ×δ Now due to Lemma 2.14 there exist λ0, ..., λ2(g+1)δ2 −1 and L0, ..., Lg SC such (g+1)δ2−1 ∈ Rj j ∈ j that λ0 >> ... >> λ(g+1)δ2−1 > 0 and L = j=0 λj(L0 + L1X1 + ... + LgXg). Now observe that we can also exchange the λ by εj. j P

Definition X.35. Let R be a real closed extension field of R and C = R[i] be the algebraic δ×δ g δ×δ R closure of R. For H (SC ) , A C and L = A + HX we define L = C g ∈ R,0 ∈g kδ ∗ D { ∈ S (k) L(C) R 0 and L = C S (k) L(C) R 0& w C 0 : w L(C)w = 0 , R,≻ | R R,0} D {R,0 ∈ | δ ∃ ∈ kδ\{ } } = . For B (k) and v = eα vα C with L(B)v = 0 set DL DL \ DL ∈ DL α=1 ⊗ ∈ M(B, v)L,R = span vα α 1, ..., δ . C{ | ∈ { }} P Lemma X.36. (Free Minkowksi theorem) Let K Sg be compact and matrix convex with kz(K)= δ < . Then K = mconv(mext(K)). ⊆ ∞ Proof. Without loss of generality suppose that 0 K. We make an induction on δ N: The case δ = 1 is just the usual Minkowski theorem∈ since the ordinary extreme points∈ of K(1) equal the matrix extreme points of mconv(K(1)). So let δ > 1 and suppose that the claim has been proven for δ 1. We conclude that K(δ 1) mconv(mext(K)). Let 2 − − ⊆ r = (g + 1)δ 1 and R be a real closed extension field of R. Take ε R> infinitesimal. − ∈ 0 Let H = mext(K). Assume that mconv(H) = K. Then we can find A1 K mconv(H). The induction start tells us that K(1) mconv6 H, in particular 0 H. By∈ Corollary\ X.34 r ⊆ ∈ we find L1 := εr(M 1 + L1X) where L1, ..., L1 Sg(δ), M 1, ..., M 1 SCδ×δ such k=0 k k 0 r ∈ 0 r ∈ that mconv(H) R,≻, A R,0. P⊆ DL1 ∈ DL1 δ2−1 r ∗ 1 Consider the continuous function γ1 : K(δ) S R+εR+...+ε R, (B, v) v L (B)v. ×r → 7→ This function attains a minimum ρ = εkλ1 0 where λ1 Rr+1. Now define 1 k=0 k ≤ ∈ λ1 L1 = L1 ρ I . We still have mconv(H) R,≻ and Z := R,0 K = . 1 δ P λ1⊗L1 1 λ1⊗L1 ◦ − R ⊆ D D ∩ 6 ∅ Additionally we achieved that K λ1⊗L1 . Z1 is compact. We know that for all B Z1 δ δ2−1⊆ D ∈ 1 there is v = α=1 eα vα such that dim M(B, v)λ ◦L1,R = dimspan v1, ..., vδ = δ ⊗ 2∈ S { } and ker(λ1 L1)(B) Cδ = span(v). If the dimension was smaller, we could compress ∗◦P ∩ B to P BP where P is the projection onto M(B, v)λ1◦L1,R and the compression would be also not in the closed matrix convex hull of the matrix extreme points (cf. Remark 5.25). This contradicts the induction hypothesis. We consider two cases: δ2−1 Case 1.1: There is a E Z1 such that mconv(E) Z1. Fix v such that 1 1 δ2 ∈ ⊇ δ ∈ S ker(λ L )(E) C = span(v) and span v1, ..., vδ = C . If E is reducible, then we know that◦ E = C∩ D and C, D mconv H{which implies} E mconv H. Therefore E is ⊕ r ∈∗ ∈ r ∗ irreducible. Suppose E = j=1 Vj AjVj for some Aj K(1), ..., K(δ) with I = j=1 Vj Vj 1 1 R ∈ ∗ 1 1 and each Vj = 0. If one (λ L )(Aj ) 0, then due to (I Vj)v = 0 also v (λ L )(E)v > 6 P◦ ≻ ⊗ 6 ◦P 0, which contradicts the choice of E. Thus all Aj are contained in Z1 = mconv(E). We saw already that E is matrix extreme in mconv(E) (Lemma 5.10). Hence all Aj are unitarily equivalent to E. This means E H. ∈ Case 1.2: There is no B Z1 such that mconv(B) Z1. Choose B Z1 and C Z1 such that C / mconv(B).∈ Set H = H B . We claim⊇ C / mconv(H∈ ). This is again∈ ∈ 1 ∪ { } ∈ 1 due to the fact that (λ1 L1)(A) R 0 for all A H. So again by Corollary X.34 we can 2 r r 2 ◦ 2 ≻ 2 2∈ g 2 2 δ×δ find L := k=0 ε (Mk + LkX) where L0, ..., Lr S (δ), M0 , ..., Mr SC such that R,≻ R,0 ∈ ∈δ2−1 mconv(H1) L2 , C L2 . The continuous function γ2 : Z1 S R + εR + r P⊆ D ∗ 2 ∈ D r k 2 × 2→ r+1 ...ε R, (D, v) v L (D)v attains a minimum ρ2 = k=0 ε λk 0 where λ R . Set 2 2 2 7→ R,≻ ≤ R,0 ∈ λ L = L ρ2Iδ. We observe that mconv(H1) λ2◦L2 and Z2 := λ2◦L2 Z1 = is ◦ − R ⊆P D D ∩ 6 ∅ compact as well as Z1 λ2◦L2 . ⊆ D 74 Again we consider the two cases 1.2 (there is B Z such that mconv(B) Z ; in ∈ 2 ⊇ 2 this case we show again that B is matrix extreme) or the case 2.2 (there is no B Z2 3 ∈ such that mconv(B) Z2; in this case we define L and Z3 like above and continue the procedure). If at some⊇ iteration m the case m.1 occurs, we are done. Assume this is not the case. Noticing that ε is infinitesimal and algebraically independent over R, we get the g r m m m 2 following: For m N let Rm = C S (δ) det k=0( λk + Mk + Lk X)(C) = 0 . Then the sequence∈ ( n R ){ ∈is a strictly| descending− sequence of varieties, which}} is m=1 m n∈N a contradiction to the Hilbert basis theorem. P T Acknowledgements First of all I want to thank my supervisor Markus Schweighofer for giving me the possibility of writing a PhD thesis, from which this paper evolved, under his supervision and helping me with mathematical suggestions, discussions and advice. I always enjoyed working in his group. I am grateful that Markus was always available for giving support and encouraged me to do research. I am indebted to Victor Vinnikov, who was present in many discussions with Markus Schweighofer and has been something like a second supervisor. His ideas and vision often proved to be very useful and lead me in the right way. Igor Klep was helping by having inspirational conversations about many different aspects of matrix convex sets with me. I am very grateful that he gave me the possibility to visit him in Ljubljana and to do research together. He made a lot of effort to bring me closer to other people working with matrix convex sets and helped me to make my work known to the mathematical community. I also appreciate having fruitful discussions with Tim Netzer, Jurij Volˇciˇc, J.W. Helton and Eric Evert.

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